transition metaltransition metaltransition metaltransition metalchemistry
TRANSCRIPT
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TRANSITION METALCHEMISTRY
2009 NEW SPECIFICATION
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CONTENTS• Definition
• Metallic properties
• Electronic configurations
• Variable oxidation state
• Coloured ions
• Complex ion formation
• Shapes of complexes
• Isomerism in complexes
• Catalytic properties
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Before you start it would be helpful to…
• Recall the definition of a co-ordinate (dative covalent) bond
H3N :H
Recall how to predict the shapes of simple molecules and ions
Transition metals
• A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.
• The highest occupied d orbital filled by an electron is d- block element
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2 Transition metals and their chemistry
A
Describe transition metals as those elements which form one or more stable ions which have incompletely filled d orbitals.
B
Derive the electronic configuration of the atoms of the d-block elements (Sc to Zn) and their simple ions from their atomic number.
CDiscuss the evidence for the electronic configurations of the elements Sc to Zn based on successive ionization energies.
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D
Recall that transition elements in general:i Show variable oxidation number in their compounds, eg redox reactions of vanadium.ii Form coloured ions in solution.iii Form complex ions involving monodentate and bidentate ligands.iv Can act as catalysts both as the elements and as their compounds.
E
Recall the shapes of complex ions limited to linear [CuCl2]-, planar [Pt(NH3)2Cl2], tetrahedral [CrCl4]-
and octahedral [Cr(NH3)6]3+, [Cu(H2O)6]2+ and other
aqua complexes.
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F
Use the chemistries of chromium and copper to illustrate and explain some properties of transition metals as follows:i. The formation of a range of compounds in which they are present in different oxidation states.ii. The presence of dative covalent bonding in complex ions, including the aqua-ion.siii. The colour or lack of colour of aqueous ions and other complex ions, resulting from the splitting of the energy levels of the d orbitals by ligands.iv. Simple ligand exchange reactions.v. Relate relative stability of complex ions to the entropy changes of ligand exchange reactions involving polydentate ligands (qualitatively only), eg EDTA.vi. Relate disproportionation reactions to standard electrode potentials and hence to Ecell.
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G
Carry out experiments to:i. Investigate ligand exchange in copper complexes.ii. Study the redox chemistry of chromium in oxidation states Cr(VI), Cr(III) and Cr(II).iii. Prepare a sample of a complex, eg chromium(II) ethanoate.
H
Recall that transition metals and their compounds are important as catalysts and that their activity may be associated with variable oxidation states of the elements or surface activity, eg catalytic converters in car exhausts.
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I
Explain why the development of new catalysts is a priority area for chemical research today and, in this context, explain how the scientific community reports and validates new discoveries and explanations, eg the development of new catalysts for making ethanoic acid from methanol and carbon monoxide with a high atom economy (green chemistry).
J
Carry out and interpret the reactions of transition metal ions with aqueous sodium hydroxide and aqueous ammonia, both in excess, limited to reactions with aqueous solutions of Cr(III), Mn(II), Fe(II), Fe(III), Ni(II), Cu(II), Zn(II).
KWrite ionic equations to show the difference between amphoteric behaviour and ligand exchange in the reactions in (J) above.
IDiscuss the uses of transition metals and/or their compounds, eg in polychromic sun glasses, chemotherapy drugs.
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Transition Metals and Living Organisms
• Iron – transport & storage of O2
• Molybdenum and Iron– Catalysts in nitrogen fixation
• Zinc – found in more than 150 bio molecules• Copper and Iron – crucial role in respiratory
cycle• Cobalt – found in vitamin B12
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Transition means “an in between state” and the transition elements come in between Group 2
and Group 3.
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Tc Ru Pd Ag CdRh
Hf Ta W Re Os Ir Au HgLa Pt
Rf Db Sg Bh Hs Mt ? ?Ac ?
H
Li
Na
K
Rb
Cs
Fr
Be
Mg
Ga Ge Se BrCa Kr
In Sn SbSr Te
Ba Tl Pb Bi Po At
Ra
Al P
N O
S Cl
F Ne
Ar
Rn
I
Si
Xe
He
B C
As
Gp 2 Gp 3
Transition Elements
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They are less reactivethan Group 1 orGroup 2 metals.
They mostly formcoloured
compounds.
They havehigh meltingtemperature.
They have high density.
Transition metalsare often referred toas ‘typical’ metals.
Mechanical properties
TransitionElements
They often act as
catalysts.
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• Similarities are more noticeable than differences although there are still some broad patterns.
• They are all dense which is what we expect of metals.
0
1
2
3
4
5
6
7
8
9
Density (g/cm-3)
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Properties – density
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Properties – melting temperature
Melting temperature show no regular pattern – other than nearly all being high which is typical of metals.(Note zinc doesn’t fit very well on either density or melting temperature.)
0200400600800
100012001400160018002000
Melting Point ( C)
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Sc Ti V Cr Mn Fe Co Ni Cu Zn
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Properties – catalysis
• A catalyst is a substance that speeds up a chemical reaction without being used up.
• Catalysts are hugely valuable in industry where they can save time and energy.
• Many transition elements ( and their compounds) are catalysts.
V2O5Ti
Used in plastic manufacture
Fe
Ni
Used in oil hydrogenation
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Transition metals are good metal
catalysts because they easily lend and
take electrons from other molecules.
A catalyst is a chemical substance that,
when added to a chemical reaction, does
not affect the thermodynamics of a
reaction but increases the rate of reaction.
Properties – catalysis
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Uses
The three most commonly known transition elements are iron or steel, copper and zinc.
iron or steel
General engineerin
g metal
copper
Electrical and plumbing
work
zinc
Galvanising steel to protect it
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Pair the metal up with its uses
iron or steel
copper
zinc
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Pair the metal catalyst with the substance.
V
Ti
Fe
Ni
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CONTENTS• Enthalpy changes
• Activation Energy
• Heterogeneous catalysis
• Specificity
• Catalytic converters
• Homogeneous catalysis
• Autocatalysis
• Enzymes
How does catalyst work?
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Before you start it would be helpful to…• know how the basics of collision theory• understand the importance of activation energy• understand the importance of increasing the rate of
reaction
CATALYSTS - background
All reactions are accompanied by changes in enthalpy.
The enthalpy rises as the reaction starts because energy is being put in to break bonds.
It reaches a maximum then starts to fall as bonds are formed and energy is released. ENTHALPY CHANGE DURING
AN EXOTHERMIC REACTION
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CATALYSTS - backgroundAll reactions are accompanied by changes in enthalpy.The enthalpy rises as the reaction starts because energy is being put in to break bonds.It reaches a maximum then starts to fall as bonds are formed and energy is released.
ENTHALPY CHANGE DURINGAN EXOTHERMIC REACTION
If the…FINAL ENTHALPY < INITIAL ENTHALPYit is anEXOTHERMIC REACTIONand ENERGY IS GIVEN OUTFINAL ENTHALPY > INITIAL ENTHALPYit is anENDOTHERMIC REACTIONand ENERGY IS TAKEN IN
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ACTIVATION ENERGY - Ea • Reactants will only be able to proceed to products if
they have enough energy
• The energy is required to overcome an energy barrier
• Only those reactants with enough energy will get over
• The minimum energy required is known as the ACTIVATION ENERGY
ACTIVATION ENERGY Ea FORAN EXOTHERMIC REACTION
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COLLISION THEORY
According to COLLISON THEORY a reaction will only take place if…
• PARTICLES COLLIDE
• PARTICLES HAVE AT LEAST A MINIMUM AMOUNT OF ENERGY
• PARTICLES ARE LINED UP CORRECTLY
To increase the chances of a successful reaction you need
to...
• HAVE MORE FREQUENT COLLISONS
• GIVE PARTICLES MORE ENERGY or
• DECREASE THE MINIMUM ENERGY REQUIRED
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NU
MB
ER
OF
MO
LE
CU
ES
W
ITH
A P
AR
TIC
UL
AR
EN
ER
GY
MOLECULAR ENERGY Ea
DUE TO THE MANY COLLISONS TAKING PLACE IN GASES, THERE IS A SPREAD OF MOLECULAR ENERGY AND VELOCITY
DUE TO THE MANY COLLISONS TAKING PLACE IN GASES, THERE IS A SPREAD OF MOLECULAR ENERGY AND VELOCITY
NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
MAXWELL-BOLTZMANN DISTRIBUTION
The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.If a catalyst is added, the Activation Energy is lowered - Ea will move to the left.
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The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.Lowering the Activation Energy, Ea, results in a greater area under the curve after Ea showing that more molecules have energies in excess of the Activation Energy
Ea
EXTRA NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
MAXWELL-BOLTZMANN DISTRIBUTION
NU
MB
ER
OF
MO
LE
CU
ES
W
ITH
A P
AR
TIC
UL
AR
EN
ER
GY
MOLECULAR ENERGY
DUE TO THE MANY COLLISONS TAKING PLACE IN GASES, THERE IS A SPREAD OF MOLECULAR ENERGY AND VELOCITY
DUE TO THE MANY COLLISONS TAKING PLACE IN GASES, THERE IS A SPREAD OF MOLECULAR ENERGY AND VELOCITY
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Catalysts work by providing…
“AN ALTERNATIVE REACTION PATHWAY WHICH HAS A LOWER ACTIVATION ENERGY”
CATALYSTS - lower Ea
A GREATER PROPORTION OF PARTICLES WILL HAVE ENERGIES
IN EXCESS OF THE MINIMUM REQUIRED SO MORE WILL REACT
WITHOUT A CATALYST WITH A CATALYST
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PRINCIPLES OF CATALYTIC ACTION
The two basic types of catalytic action are …
HETEROGENEOUS CATALYSIS
and
HOMOGENEOUS CATALYSIS
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Format Catalysts are in a different phase to the reactants e.g. a solid catalyst in a gaseous reaction
Action takes place at active sites on the surface of a solid gases are adsorbed onto the surface they form weak bonds with metal atoms
Catalysis is thought to work in three stages...
Adsorption
Reaction
Desorption
Heterogeneous Catalysis
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Heterogeneous Catalysis
For an explanation of what happens click on the numbers in turn, starting with
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Heterogeneous Catalysis
Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.
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Heterogeneous Catalysis
Desorption (STEP 4)There is a re-arrangement of electrons and the products are then released from the active sites
Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.
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ANIMATION
Heterogeneous Catalysis
Desorption (STEP 4)There is a re-arrangement of electrons and the products are then released from the active sites
Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.
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Heterogeneous Catalysis
ANIMATION
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STRENGTH OF ADSORPTION
The STRENGTH OF ADSORPTION is critical ...
too weak Ag little adsorption - few available d orbitalstoo strong W molecules remain on the surface preventing further reactionjust right Ni/Pt molecules are held but not too strongly so they can get away
Catalysis of gaseous reactions can lead to an increase in rate in several ways
• one species is adsorbed onto the surface and is more likely to undergo a collision
• one species is held in a favourable position for reaction to occur• adsorption onto the surface allows bonds to break and fragments react
quicker• two reactants are adsorbed alongside each other give a greater
concentration
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[Ar]
Cu [Ar] 3d10 4s1 4p0 4d0
How does catalyst work?
C2H5OH CH3CHO + H2
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Specificity
In some cases the choice of catalyst can influence the productsEthanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened.
CLICK HERE FOR ANIMATION
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SpecificityIn some cases the choice of catalyst can influence the products
Ethanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened.
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SpecificityIn some cases the choice of catalyst can influence the products
C2H5OH CH3CHO + H2 C2H5OH C2H4 + H2O
Ethanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened. Alumina DehydrationCopper Dehydrogenation (oxidation)
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Ethanol undergoes two different reactions depending on the metal used as the catalyst.
COPPER Dehydrogenation (oxidation)
C2H5OH CH3CHO + H2
The active sites are the same distanceapart as the length of an O-H bond
It breaks to release hydrogen
ALUMINA Dehydration (removal of water)
C2H5OH C2H4 + H2O
The active sites are the same distanceapart as the length of a C-O bond
It breaks to release an OH group
Specificity
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Poisoning
Impurities in a reaction mixture can also adsorb onto the surface of a catalyst thus removing potential sites for gas molecules and decreasing efficiency.
expensive because... the catalyst has to replaced the process has to be shut
down
examples Sulphur Haber processLead catalytic converters in cars
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Catalytic convertersPURPOSE removing the pollutant gases formed in
internal combustion enginesPOLLUTANTS CARBON MONOXIDE
NITROGEN OXIDESUNBURNT HYDROCARBONS
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Catalytic convertersPURPOSE removing the pollutant gases formed
in internal combustion enginesPOLLUTANTS CARBON MONOXIDE
NITROGEN OXIDESUNBURNT HYDROCARBONS
CONSTRUCTION made from alloys of platinum, rhodium and palladium catalyst is mounted in a support medium to spread it out honeycomb construction to ensure maximum gas contact finely divided to increase surface area / get more collisions
involves HETEROGENEOUS CATALYSIS
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CO + Unburned Hydrocarbons + O2 CO2 + H2O
2NO + 2NO2 2N2 + 3O2
Catalytic converters
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Pollutant gases
Carbon monoxide CO
Originincomplete combustion of hydrocarbons in petrol when not enough oxygen is present to convert all the carbon to carbon dioxide
C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)
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Pollutant gases
Carbon monoxide CO
Origin incomplete combustion of hydrocarbons in petrol when not enough oxygen is present to convert all the carbon to carbon dioxide
C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)
Effect poisonouscombines with haemoglobin in bloodprevents oxygen being carried to cells
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Pollutant gases
Carbon monoxide CO
Origin incomplete combustion of hydrocarbons in petrol when not enough
oxygen is present to convert all the carbon to carbon dioxide
C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)
Effect poisonouscombines with haemoglobin in bloodprevents oxygen being carried to cells
Removal 2CO(g) + O2(g) 2CO2(g)
2CO(g) + 2NO(g) N2(g) + 2CO2(g)
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Pollutant gases
Oxides of nitrogen NOx - NO, N2O and NO2
Origin nitrogen and oxygen combine under high temperature conditions
nitrogen combines with oxygen N2(g) + O2(g) 2NO(g)
nitrogen monoxide is oxidised 2NO(g) + O2(g) 2NO2(g)
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Pollutant gases
Oxides of nitrogen NOx - NO, N2O and NO2
Origin nitrogen and oxygen combine under high temperature conditions
nitrogen combines with oxygen N2(g) + O2(g) 2NO(g)
nitrogen monoxide is oxidised 2NO(g) + O2(g) 2NO2(g)
Effect photochemical smog - irritating to eyes, nose and throatproduces low level ozone - affects plant growth
- is irritating to eyes, nose and throat
i) sunlight breaks down NO2 NO2 NO + O
ii) ozone is produced O + O2 O3
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Pollutant gases
Oxides of nitrogen NOx - NO, N2O and NO2
Origin nitrogen and oxygen combine under high temperature conditions
nitrogen combines with oxygen N2(g) + O2(g) 2NO(g)
nitrogen monoxide is oxidised 2NO(g) + O2(g) 2NO2(g)
Effect photochemical smog - irritating to eyes, nose and throatproduces low level ozone - affects plant growth
- is irritating to eyes, nose and throat
i) sunlight breaks down NO2 NO2 NO + O
ii) ozone is produced O + O2 O3
Removal 2CO(g) + 2NO(g) N2(g) + 2CO2(g)
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Pollutant gases
Unburnt hydrocarbons CxHy
Origin insufficient oxygen for complete combustion
Effect toxic and carcinogenic (causes cancer)
Removal catalyst aids complete combustion
C8H18(g) + 12½O2(g) 8CO2(g) + 9H2O(l)
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Homogeneous Catalysis
Action • catalyst and reactants are in the same phase
• reaction proceeds through an intermediate species of lower energy
• there is usually more than one reaction step
• transition metal ions are often involved - oxidation state changes
Example
Acids Esterificaton
Conc. H2SO4 catalyses the reaction between acids and alcohols
CH3COOH + C2H5OH CH3COOC2H5 + H2O
NB Catalysts have NO EFFECT ON THE POSITION OF EQUILIBRIUM
but they do affect the rate at which equilibrium is reached
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Homogeneous Catalysis
Action • catalyst and reactants are in the same phase
• reaction proceeds through an intermediate species of lower energy
• there is usually more than one reaction step
• transition metal ions are often involved - oxidation state changes
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Homogeneous Catalysis
Action • catalyst and reactants are in the same phase
• reaction proceeds through an intermediate species with of energy
• there is usually more than one reaction step
• transition metal ions are often involved - oxidation state changes
Examples
Gases Atmospheric OZONE breaks down naturally O3 O• + O2
- it breaks down more easily in the presence of chlorofluorocarbons (CFC's).
There is a series of complex reactions but the basic process is :-
CFC's break down in the presence ofUV light to form chlorine radicals CCl2F2 Cl• +
• CClF2
chlorine radicals then react with ozone O3 + Cl• ClO• + O2
chlorine radicals are regenerated ClO• + O O2 + Cl•
Overall, chlorine radicals are not used up so a small amount of CFC's can
destroy thousands of ozone molecules before the termination stage.
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Transition metal compoundsThese work because of their ability to change oxidation state
1. Reaction between iron(III) and vanadium(III)
The reaction is catalysed by Cu2+
step 1 Cu2+ + V3+ Cu+ + V4+
step 2 Fe3+ + Cu+ Fe2+ + Cu2+
overall Fe3+ + V3+ Fe2+ + V4+
1. Reaction between iron(III) and vanadium(III)
The reaction is catalysed by Cu2+
step 1 Cu2+ + V3+ Cu+ + V4+
step 2 Fe3+ + Cu+ Fe2+ + Cu2+
overall Fe3+ + V3+ Fe2+ + V4+
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Transition metal compoundsThese work because of their ability to change oxidation state
2. Reaction between I¯ and S2O82-
A slow reaction because REACTANTS ARE NEGATIVE IONS REPULSION
Addition of iron(II) catalyses the reaction
step 1 S2O82- + 2Fe2+ 2SO4
2- + 2Fe3+
step 2 2Fe3+ + 2I¯ 2Fe2+ + I2
overall S2O82- + 2I¯ 2SO4
2- + I2
2. Reaction between I¯ and S2O82-
A slow reaction because REACTANTS ARE NEGATIVE IONS REPULSION
Addition of iron(II) catalyses the reaction
step 1 S2O82- + 2Fe2+ 2SO4
2- + 2Fe3+
step 2 2Fe3+ + 2I¯ 2Fe2+ + I2
overall S2O82- + 2I¯ 2SO4
2- + I2
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Auto-catalysis
Occurs when a product of the reaction catalyses the reaction itself
It is found in the reactions of manganate(VII) with ethandioate
2MnO4¯ + 16H+ + 5C2O42- 2Mn2+ + 8H2O + 10CO2
The titration needs to be carried out at 70°C because the reaction is slow as Mn2+ is formed the reaction speeds up; the Mn2+ formed acts as the catalyst
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A redox titration.
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Activity is affected by ...
temperature - it increases until the protein is denatured
substrate concentration - reaches a maximum when all sites are blocked
pH- many catalysts are amino acids which can be protenated
being poisoned - when the active sites become “clogged” with unwanted
ENZYMES
Action enzymes are extremely effective biologically active catalysts they are homogeneous catalysts, reacting in solution with body fluids only one type of molecule will fit the active site “lock and key” mechanism makes enzymes very specific as to what they catalyse.
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ENZYMESAction enzymes are extremely effective biologically active catalysts they are homogeneous catalysts, reacting in solution with body fluids only one type of molecule will fit the active site “lock and key” mechanism makes enzymes very specific as to what they catalyse.
A B C
A Only species with the correct shape can enter the active site in the enzymeB Once in position, the substrate can react with a lower activation energyC The new products do not have the correct shape to fit so the complex breaks up
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ENZYMES
ANIMATED ACTION
A Only species with the correct shape can enter the active site in the enzymeB Once in position, the substrate can react with a lower activation energyC The new products do not have the correct shape to fit so the complex breaks up
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Introduction
• d-block elements
locate between the s-block andp-block
known as transition elements
occur in the fourth and subsequent periods of the Periodic Table
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period 4
period 5
period 6
period 7
d-block elements
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Introduction
Transition elements are elements that contain an incomplete d sub-shell (i.e. d1 to d9) in at least one of the oxidation states of their compounds.
3d0
3d10
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Sc and Zn are not transition elements because
They form compounds with only one oxidation state in which the d sub-shell are NOT imcomplete.
Sc Sc3+ 3d0 Zn Zn2+ 3d10
Introduction
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Cu
Cu+ 3d10 not transitional
Cu2+ 3d9 transitional
Introduction
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The first transition series
the first horizontal row of the d-block elements
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The building up of electronic configurations of elements follow:
Aufbau principle
Pauli exclusion principle
Hund’s rule
Electronic Configurations
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Vanadium is classified as a transition metal. This is because vanadium
A is a d-block element.B has incompletely filled d orbitals.C forms stable ions with incompletely filled d orbitals.D forms stable ions in which it has different oxidation states. Shafeeu
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Aufbau Principle: – orbitals fill in order of increasing energy from
lowest energy to highest energyPauli Exclusion Principle:
– only two electrons can occupy an orbital and their spins must be paired
Hund’s Rule: – when orbitals of equal energy are available but
there are not enough electrons to fill all of them, one electron is added to each orbital before a second electron is added to any one of them
Electronic Configurations
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K 1s2 2s2 2p6 3s2 3p6 4s1
Ca 1s2 2s2 2p6 3s2 3p6 4s2
Sc 1s2 2s2 2p6 3s2 3p6 4s2 3d1
Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2
V 1s2 2s2 2p6 3s2 3p6 4s2 3d3
Cr 1s2 2s2 2p6 3s2 3p6 4s1 3d5
Mn 1s2 2s2 2p6 3s2 3p6 4s2 3d5
Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Co 1s2 2s2 2p6 3s2 3p6 4s2 3d7
Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8
Cu 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Zn 1s2 2s2 2p6 3s2 3p6 4s2 3d10
ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
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Element Atomic number
Electronic configuration
ScandiumTitaniumVanadiumChromiumManganeseIronCobaltNickelCopperZinc
21222324252627282930
[Ar]3d14s2
[Ar]3d24s2
[Ar]3d34s2
[Ar]3d54s1
[Ar]3d54s2
[Ar]3d64s2
[Ar]3d74s2
[Ar]3d84s2
[Ar]3d104s1
[Ar]3d104s2
Electronic configurations of the first series of d-block elements
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Relative energy levels of orbitals before and after filling with electrons
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d-Block elements (transition elements):• Lie between s-block and p-block elements • Occur in the fourth and subsequent periods• All contains incomplete d sub-shell (i.e. 1 – 9
electrons) in at least one of their oxidation state
Scandium
Titanium
Vanadium Chromium
Manganese
ZincCopperNickel
CobaltIron
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Strictly speaking, scandium (Sc) and zinc
(Zn) are not transitions elements
∵ Sc forms Sc3+ ion which has an empty
d sub-shell (3d0)
Zn forms Zn2+ ion which has a
completely filled d sub- shell (3d10)
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Cu+ is not a transition metal ion as it has a completely filled d sub-shell
Cu2+ is a transition metal ion as it has an incompletely filled d sub-shell
Cu shows some intermediate behaviour between
transition and non-transition elements because of
two oxidation states, Cu(I) & Cu(II)
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THE FIRST ROW TRANSITION ELEMENTS
Definition Transition elements forming one or more stable ions with partially filled (incomplete) d-sub shells.
The first row runs from scandium to zinc filling the 3d orbitals.
Properties arise from an incomplete d sub-shell in atoms or ionsMetallic all the transition elements are metals properties strong metallic bonds due to small ionic size and close packing higher melting, boiling points and densities than s-block metals
K Ca Sc Ti V Cr Mn Fe Co
T m/ °C 63 850 1400 1677 1917 1903 1244 1539 1495densityg cm-3 0.86 1.55 3 4.5 6.1 7.2 7.4 7.9 8.9
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4s
3 3p
3d
44p
4d
4f
ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
POTASSIUM
1s2 2s2 2p6 3s2 3p6 4s1
In numerical terms one would expect the 3d orbitals to be filled next.
However, because the principal energy levels get closer together as you go further from the nucleus coupled with the splitting into sub energy levels, the 4s orbital is of a LOWER ENERGY than the 3d orbitals so gets filled first.
‘Aufbau’ Principle‘Aufbau’ Principle
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4s
3 3p
3d
44p
4d
4f
CALCIUM
1s2 2s2 2p6 3s2 3p6 4s2
As expected, the next electron in pairs up to complete a filled 4s orbital.
This explanation, using sub levels fits in with the position of potassium and calcium in the Periodic Table. All elements with an -s1 electronic configuration are in Group I and all with an -s2 configuration are in Group II.
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4s
3 3p
3d
44p
4d
4f
SCANDIUM
1s2 2s2 2p6 3s2 3p6 4s2 3d1
With the lower energy 4s orbital filled, the next electrons can now fill p the 3d orbitals. There are five d orbitals. They are filled according to Hund’s Rule.
BUT WATCH OUT FOR TWO SPECIAL CASES.
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4s
3 3p
3d
44p
4d
4f
TITANIUM
1s2 2s2 2p6 3s2 3p6 4s2 3d2
The 3d orbitals are filled according to Hund’s rule so the next electron doesn’t pair up but goes into an empty orbital in the same sub level.
HUND’S RULE OFMAXIMUM
MULTIPLICITY
HUND’S RULE OFMAXIMUM
MULTIPLICITY
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4s
3 3p
3d
44p
4d
4fVANADIUM
1s2 2s2 2p6 3s2 3p6 4s2 3d3
The 3d orbitals are filled according to Hund’s rule so the next electron doesn’t pair up but goes into an empty orbital in the same sub level.
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HUND’S RULE OFMAXIMUM
MULTIPLICITY
HUND’S RULE OFMAXIMUM
MULTIPLICITY
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4s
3 3p
3d
44p
4d
4fCHROMIUM
1s2 2s2 2p6 3s2 3p6 4s1 3d5
One would expect the configuration of chromium atoms to end in 4s2 3d4.
To achieve a more stable arrangement of lower energy, one of the 4s electrons is promoted into the 3d to give six unpaired electrons with lower repulsion.
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4s
3 3p
3d
44p
4d
4f
MANGANESE
1s2 2s2 2p6 3s2 3p6 4s2 3d5
The new electron goes into the 4s to restore its filled state.
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4s
3 3p
3d
44p
4d
4f
MANGANESE
1s2 2s2 2p6 3s2 3p6 4s2 3d5
The new electron goes into the 4s to restore its filled state.
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4s
3 3p
3d
44p
4d
4f
IRON
1s2 2s2 2p6 3s2 3p6 4s2 3d6
Orbitals are filled according to Hund’s Rule. They continue to pair up.
HUND’S RULE OFMAXIMUM MULTIPLICITY
HUND’S RULE OFMAXIMUM MULTIPLICITY
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4s
3 3p
3d
44p
4d
4f
COBALT
1s2 2s2 2p6 3s2 3p6 4s2 3d7
Orbitals are filled according to Hund’s Rule. They continue to pair up.
HUND’S RULE OFMAXIMUM MULTIPLICITY
HUND’S RULE OFMAXIMUM MULTIPLICITY
Anhydrous
Hexahydrate
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ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
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4s
3 3p
3d
44p
4d
4f
NICKEL
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Orbitals are filled according to Hund’s Rule. They continue to pair up.
HUND’S RULE OFMAXIMUM MULTIPLICITY
HUND’S RULE OFMAXIMUM MULTIPLICITY
INC
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ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
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4s
3 3p
3d
44p
4d
4f
COPPER
1s2 2s2 2p6 3s2 3p6 4s1 3d10
One would expect the configuration of copper atoms to end in 4s2 3d9.
To achieve a more stable arrangement of lower energy, one of the 4s electrons is promoted into the 3d.
HUND’S RULE OFMAXIMUM MULTIPLICITY
HUND’S RULE OFMAXIMUM MULTIPLICITY
INC
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ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
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4s
3 3p
3d
44p
4d
4fZINC
1s2 2s2 2p6 3s2 3p6 4s2 3d10
The electron goes into the 4s to restore its filled state and complete the 3d and 4s orbital filling.
INC
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ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS
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Electronic Electronic ConfigurationElement Z 3d 4sSc 21 [Ar]
Ti 22 [Ar]
V 23 [Ar]
Cr 24 [Ar]
Mn 25 [Ar]
Fe 26 [Ar]
Co 27 [Ar]
Ni 28 [Ar]
Cu 29 [Ar]
Zn 30 [Ar]
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VARIABLE OXIDATION STATES
Arises from the similar energies required for removal of 4s and 3d electrons
Maximum rises across row to manganese
Maximum falls as the energy required to remove more
electrons becomes very high all (except scandium)
have an M2+ ion stability of +2 state increases across
the row due to increase in the 3rd Ionisation Energy
THE MOST IMPORTANT STATES ARE IN RED
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Oxidation states of the elements of the first transition series in their compounds
Element Possible oxidation state
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Element Possible oxidation state
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
+3
+1 +2 +3 +4
+1 +2 +3 +4 +5
+1 +2 +3 +4 +5+6
+1 +2 +3 +4 +5+6 +7
+1 +2 +3 +4 +5+6
+1 +2 +3 +4 +5
+1 +2 +3 +4 +5
+1 +2 +3
+2
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Ti Sc V Cr Mn Fe Co Ni Cu Zn
+1+2
+3+4
+5+6
+2 +2 +2 +2 +2 +2 +2 +2
+6+7
+3+3 +3+4
+3+4
+3 +3 +3+4 +4 +4 +4+5 +5 +5 +5
+6
When electrons are removed they come from the 4s orbitals first Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2
Ti2+ 1s2 2s2 2p6 3s2 3p6 3d2
Ti3+ 1s2 2s 2p6 3s2 3p6 3d1
Ti4+ 1s2 2s2 2p6 3s2 3p6
Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Cu+ 1s2 2s2 2p6 3s2 3p6 3d10
Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9
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Oxidation states
Oxides / Chloride
+1Cu2O
Cu2Cl2
+2
TiO VO CrO MnO FeO CoO NiOCuO ZnO
TiCl2 VCl2 CrCl2 MnCl2 FeCl2 CoCl2 NiCl2CuCl2 ZnCl2
+3
Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Ni2O3 •
xH2O
ScCl3 TiCl3 VCl3 CrCl3 MnCl3 FeCl3
+4TiO2 VO2 MnO2
TiCl4 VCl4 CrCl4
+5 V2O5
+6 CrO3
+7 Mn2O7
Oxidation states of the elements of the first transition series in their oxides and chlorides
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The element zinc, with electronic configuration
1s22s22p63s23p63d104s2, is not regarded as a
transition element because
A the oxide of zinc is amphoteric.
B none of its ions has an unpaired electron in the
d -sub shell.
C it does not readily form complex ions.
D it has a boiling temperature low enough for it to be
easily distilled. Shahudaan
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Ioniz
atio
n E
nth
alp
y
Element
Ionization enthalpy (kJ mol–1)
1st 2nd 3rd 4th
KCa
418590
3 0701 150
4 6004 940
5 8606 480
ScTiVCrMnFeCoNiCuZn
632661648653716762757736745908
1 2401 3101 3701 5901 5101 5601 6401 7501 9601 730
2 3902 7202 8702 9903 2502 9603 2303 3903 5503 828
7 1104 1704 6004 7705 1905 4005 1005 4005 6905 980
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20 21 22 23 24 25 26 27 28 29 30 31 32
Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn 31 32
4000
3500
3000
2500
2000
1500
1000
Successive ionization energy data for evidence for electronic configuration th
ird
second
first
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• Some abnormal high ionization enthalpy, e.g. 1st I.E. of Zn, 2nd I.E. of Cr & Cu and the 3rd I.E. of Mn
∵The removal of an electron from a fully-filled or half-filled sub-shell requires a relatively large amount of energy
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7000
6000
5000
4000
3000
2000
1000
0 Sc Ti V Cr Mn Fe Co Ni Cu Zn 3d1 3d2 3d3 3d5 3d5 3d6 3d7 3d8 3d10 3d10 4s2 4s2 4s2 4s1 4s2 4s2 4s2 4s2 4s1 4s2
3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6
Ion
izat
ion
en
erg
y (
kJ /
mo
l)
second
third
first
fourth
Successive ionization energy data for evidence for electronic configuration
Electron has to be removed from completely filled 3p sub shell
3d5
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Which of the following successive ionization
energies (values in kJ mol-1) could have come
from a transition element?
A 496 4563 6913 9544 13352 16611 20115 25941
B 590 1145 4912 6474 8144 10496 12320 14207
C 717 1509 3249 4940 6985 9200 11508 18956
D 2081 3952 6122 9370 12177 15239 19999 23069
Nuha
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Ionization Enthalpy
ElementIonization enthalpy (kJ mol–1)
1st 2nd 3rd 4th
KCa
418590
3 0701 150
4 6004 940
5 8606 480
ScTiVCrMnFeCoNiCuZn
632661648653716762757736745908
1 2401 3101 3701 5901 5101 5601 6401 7501 9601 730
2 3902 7202 8702 9903 2502 9603 2303 3903 5503 828
7 1104 1704 6004 7705 1905 4005 1005 4005 6905 980
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Explain the following variation in terms of electronic configurations.
(a) The second ionization enthalpies of both Cr and Cu are higher than those of their next elements respectively.
Answer(a)The second ionization enthalpies of both Cr and Cu are higher
than those of their next elements respectively. In the case of Cr, the second ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell, which has extra stability. Therefore, this second ionization enthalpy is relatively high. The case is similar for copper where its second ionization enthalpy involves the removal of an electron from a fully-filled 3d sub-shell which also has extra stability. Thus, its second ionization enthalpy is also relatively high.
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Explain the following variation in terms of electronic configurations. Happiness
(b)The third ionization enthalpy of Mn is higher than that of its next element.
Answer(b) The third ionization enthalpy of Mn is higher than that of its next element. It is because its third ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell which has extra stability. Therefore, its third ionization enthalpy is relatively high.
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Zn
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Oxidation States of the Transition MetalsSome oxidation states, however, are more common than others. The most common oxidation states of the first series of transition metals are given in the table below. Some of these oxidation states are common because theyare relatively stable. Others describe compounds that are not necessarily stable but which react slowly. Possible electron configurations of transition metal ionsSc3+, [Ar]3d0
Ti2+, [Ar]; Ti4+, [Ar]3d0
V2+, [Ar]3d3; V3+, [Ar]3d2; V4+, [Ar]3d1; V5+, [Ar]3d0
Cr2+, [Ar]3d4; Cr3+, [Ar]3d3; Cr4+, [Ar]3d2; Cr6+, [Ar]Mn2+,[Ar]3d5; Mn4+, [Ar]3d3; Mn7+, [Ar] 3d0
Fe2+, [Ar]3d6; Fe3+, [Ar]3d5
Co2+, [Ar]3d7; Co3+, [Ar]3d6
Ni2+, [Ar]3d8
Cu+, [Ar]3d10; Cu2+, [Ar]3d9
Zn2+, [Ar]3d10
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Common Oxidation States of the First Series of Transition Metals
One point about the oxidation states of transition metals deserves particular attention: Transition-metal ions with charges larger than +3 cannot exist in aqueous solution.
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COLOURED IONS
Theory ions with a d10 (full) or d0 (empty) configuration
are colourle ions with partially filled d-orbitals
tend to be coloured it is caused by the ease of
transition of electrons between energy levels
energy is absorbed when an electron is promoted
to a higher level the frequency of light is
proportional to the energy difference
ions with d10 (full) Cu+,Ag+ Zn2+
or d0 (empty) Sc3+ configuration are colourlesse.g. titanium(IV) oxide TiO2 is white
colour depends on ...transition element oxidation state ligand , coordination number
A characteristic of transition metals is their ability to form coloured compounds
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A hydrated transition metal ion is colourless.
Which of the following could be the electronic
configuration of this ion?
A [Ar] 3d54s2
B [Ar] 3d8
C [Ar] 3d104s2
D [Ar] 3d100
Ayaman
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Cu2+ [Ar] 3d9 4s0 4p0 4d0
[Ar]
[Ar]
Cu2+ [Ar] 3d9 4s2 4p6 4d4
How does empty orbitals work?
[Cu(H2O)6]2+
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Cu2+
Dative Covalent Bond- Co-ordination bond
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Reagent\Ion and initial
colour
Cr3+(aq)
green
Mn2+(aq) very
pale pink ~ colourless
Fe2+(aq)
pale green
Fe3+(aq)
yellow-brown
Co2+(aq)
pink
Ni2+(aq)
green
Cu2+(aq)
blue
Zn2+(aq)
colourless
Al3+(aq)
colourless
initial NaOH(aq)
strong base/alkali
green gel. ppt. of
Cr(OH)3
white gel. ppt. but
darkens with oxidation
Mn(OH)2
Mn2O3
MnO2
dark green ppt. =>
brown on oxidation
Fe(OH)2
Fe(OH)3
brown gel. ppt. of
Fe(OH)3
blue gel. ppt. that
turns pink on standing
Co(OH)2
green gel. ppt. of
Ni(OH)2
gel. blue ppt. of
Cu(OH)2
white gel. ppt. of
Zn(OH)2
white gel. ppt. of
Al(OH)3
excess NaOH(aq)
strong base/alkali
ppt. dissolves to give clear
green solution of
complex ion
[Cr(OH)6]3-
no further effect - just as above with more oxidation
no further effect - just as above with more oxidation
no further effect - just as above with more oxidation
no further effect
no further effect
no further effect ppt. dissolves -
clear solution,
colourless complex ion
[Zn(OH)4]2-
ppt. dissolves -
clear solution,
colourless complex ion
[Al(OH)6]3-
initial NH3(aq)
weak base/alkali
green gel. ppt. of
Cr(OH)3
white ppt. darkens with
oxidation from O2
Mn(OH)2
Mn2O3
MnO2
dark green ppt. turns brown -
oxidation
Fe(OH)2
Fe(OH)3
brown gel. ppt. of
Fe(OH)3
blue gel. ppt. that turns pink on
standing
Co(OH)2
green gel. ppt. of
Ni(OH)2
gel. blue ppt. of
Cu(OH)2
white gel. ppt. of
Zn(OH)2
white gel. ppt. of
Al(OH)3
excess NH3(aq)
weak base/alkali
dissolves - clear green solution of
complex ion
[Cr(NH3)6]3+
no further effect
no further effect
no further effect
ppt. dissolves - clear brown solution of
complex ion
[Co(NH3)6]2+
dissolves - clear pale
blue solution of complex
ion
[Ni(NH3)6]2+
ppt. dissolves to give clear blue
solution of complex ion
[Cu(NH3)4(H2O)2]2+
ppt. dissolves -
clear colourless solution of
[Zn(NH3)4]2+
no further effect
Colours of complex ions
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Transition metal ion solution
Addition of sodium hydroxide or ammonia solution
Proton transfer
Addition of sodium hydroxide solution
Precipitate dissolves Precipitate dissolves
Ligand exchange Proton transfer
Addition of ammonia solution
Precipitate forms
Ni2+ , Cu2+, Zn2+ ,Cr3+ ,Co2+ Cr3+ , Zn2+
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[Cr(H2O)6]3+ (aq) [Fe(H2O)6]
2+ (aq) [Co(H2O)6]2+ (aq)
[Cu(H2O)6]2+ (aq)[Cu(H2O)6]
2+ (aq)
[Mn(H2O)6]2+ (aq)
Chemistry ofNi2+ , Cu2+,
Zn2+ Cr3+ ,Co2+
[Fe(H2O)6]3+(aq)
[Zn(NH3)4] 2+(aq)
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Aqueous metal ion reactions
Fe2+
Co2+
Cu2+
Fe3+
Cr3+
Zn2+
Fe2+(aq) ligand substitution
Cu2+(aq) ligand substitution
Cr3+(aq) acidity reaction
Zn2+(aq) acidity reaction
reactions photos
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Mn2+
Ni2+
Co2+(aq) reaction
Fe3+(aq) ligand substitution
Ni2+(aq) acidity reaction
Mn2+(aq) acidity reaction
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Fe2+ iron(II)
Add
NaOH (aq) (little)
[Fe(H2O)6]2+ (aq) green solution
Fe(H2O)4(OH)2 (s) green ppt(turns brown in air)
NaOH (aq) (excess)
NH3 (aq) (little)
NH3 (aq) (excess)
CO32- (aq)
FeCO3 (s)
green ppt
Does not dissolve
precipitation
photoligand substitution
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Co2+ cobalt(II)Add
NaOH (aq) (little)
[Co(H2O)6]2+ (aq) pink solution
Co(H2O)4(OH)2 (s) blue ppt(turns pink slowly and then brown in air)NaOH (aq) (excess)
NH3 (aq) (little)
NH3 (aq) (excess)[Co(NH3)6]
2+ (aq)
CO32- (aq)
CoCO3 (s) pink ppt
pale-yellow solution(turns brown in air)
Cl- (aq) (conc) [CoCl4]2- (aq) blue solution
acidity reactions
ligand substitution
precipitation
ligand substitution
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Cu2+ copper(II)Add
NaOH (aq) (little)
[Cu(H2O)6]2+ (aq) blue solution
Cu(H2O)4(OH)2] (s)
blue precipitateNaOH (aq) (excess)
NH3 (aq) (little)
NH3 (aq) (excess) [Cu(NH3)4(H2O) 2]2+ (aq)
CO32- (aq) CuCO3 (s) blue precipitate
deep blue solution
Cl- (aq) (conc) [CuCl4]2- (aq) green solution
acidity reactions
ligand substitution
precipitation
ligand substitution
photo
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acidity reactions
Fe3+ iron(III)
Add
NaOH (aq) (little)
[Fe(H2O)6]3+ (aq) violet solution
Fe(H2O)3(OH)3 (s)
brown precipitateNaOH (aq) (excess)
NH3 (aq) (little)
CO32- (aq)
Fe(H2O)3(OH)3 (s)
brown precipitate andcolourless bubbles of CO2 (g)
acidity reaction
NH3 (aq) (excess)
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acidity reaction
purple solution
acidity reaction
ligand substitution
acidity reactions
Cr3+ chromium(III)Add
NaOH (aq) (little)
[Cr(H2O)6]3+ (aq) red-blue solution
Cr(H2O)3(OH)3 (s)
green precipitateNH3 (aq) (little)
NaOH (aq) (excess) [Cr(OH)6]3- (aq)
NH3 (aq) (excess) [Cr(NH3)6]3+ (aq)
green solution
CO32- (aq)
Cr(H2O)3(OH)3 (s)
green precipitate andcolourless bubbles of CO2 (g)
photo
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acidity reactions
Zn2+ zinc(II)
Add
NaOH (aq) (little)
Zn2+ (aq) colourless solution
[Zn(OH)2 (H2O)2](s)
precipitateNH3 (aq) (little)
NaOH (aq) (excess) [Zn(OH)4]2- (aq)
colourless solution
NH3 (aq) (excess)
photo
[Zn(NH3)4] 2+(aq)
white
Ligand substitution
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Photographs
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Fe2+(aq) ligand substitution
Cu2+(aq) acidity reaction reaction
Cr3+(aq) acidity reaction
Zn2+(aq) acidity reaction
Co2+(aq) acidity reaction
Fe3+(aq) acidity reaction
Ni2+(aq) acidity reaction
Mn2+(aq) acidity reaction
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Fe2+ iron(II) photo
[Fe(H2O)6]2+ (aq)
green solution
Fe(H2O)4(OH)2 (s)
green precipitate[turns brown in air) insoluble in excess NaOH(aq) or NH3(aq)]
NaOH(aq) or NH3(aq)
Ligand substitution
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Cu2+ copper(II)
[Cu(H2O)6]2+ (aq)
blue solution
Cu(H2O)4(OH)2] (s)
NaOH(aq) or NH3(aq)
blue precipitate soluble in excessNH3(aq)
acidity reaction Click me.I’ll
show you?
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Cr3+ chromium(III) photos
[Cr(H2O)6]3+ (aq)
red-blue solutionappears green because of hydrolysis
Cr(H2O)3(OH)3 (s)green precipitate
acidity reactions
NaOH (aq) or NH3 (aq)
Click for more …
excess NaOH (aq)
[Cr(OH)6]3- (aq)
green solution
Click me. I’ll show you?
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Cr(H2O)3(OH)3 (s) + NH3
green precipitate
[Cr(NH3)6]3+(aq)
Yellowish
Violet solution
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Zn2+ zinc(ii) photo
[Zn(H2O)4]2+ (aq)
colourless solution
[Zn(OH)2 (H2O)2](s)
white precipitate
acidity reactions
Click for more …excess NaOH (aq)
NaOH (aq) or NH3 (aq)
[Zn(OH)2 (OH)4]2-(aq)
colourless solution
[Zn(OH)2 (H2O)2](s)
[Zn(NH3)4] 2+(aq)
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acidity reaction
purple solution
acidity reaction
ligand substitution
acidity reactions
Ni2+ nickel(ii)Add
NaOH (aq) (little)
[Cr(H2O)6]3+ (aq) red-blue solution
Cr(H2O)3(OH)3 (s) green ppt
NH3 (aq) (little)
NaOH (aq) (excess) [Cr(OH)6]3- (aq)
NH3 (aq) (excess) [Cr(NH3)6]3+ (aq)
green solution
CO32- (aq) Cr(H2O)3(OH)3 (s) green ppt
colourless bubbles of CO2 (g)
photo
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Mn2+ manganese(ii)Add
NaOH (aq) (little)
[Mn(H2O)6]2+ (aq) red-blue solution
Mn(H2O)3(OH)3 (s) buff ppt
NH3 (aq) (little)
NaOH (aq) (excess)
NH3 (aq) (excess)
CO32- (aq)
Mn(H2O)3(OH)3 (s) green ppt
colourless bubbles of CO2 (g)
photo
Insoluble
Insoluble
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Mn2+(aq) + 2NH3(aq) + 2H2O(l)
Mn(OH)2(s)+2NH4+(aq)
Mn2+(aq) + 2OH-(aq) Mn(OH)2(s)
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Ni(H2O)6]2+(aq) + 2-OH (aq)
Ni(OH)2 (H2O)4](s) + 2H2O
Pale green ppt soluble in excess ammonia
Ni2+(aq) + 2-OH(aq) Ni(OH)2(s)
Click me I’ll show you?
Nice?
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Ni(OH)2 (H2O)4](s) + 6NH3(aq)
[Ni(NH3)6]2+(aq) + 2-OH+4H2Olavender blue
Oh?
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[Fe(H2O)6]2+(aq) + 3-OH (aq)
[Fe(OH)3(H2O)3](s) +3H2O red brown ppt
Insoluble in excess NaOH and NH3
With NaOH and NH3
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[Co(H2O)6]2+(aq) + 2-OH (aq)
[Co(OH)2 (H2O)4](s) Blue ppt
Soluble excess ammonia
NaOH or NH3
Click me. I’ll show you?
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[Co(NH3)6]2+(aq) +4H2O +2-OH (aq)
[Co(OH)2 (H2O)4](s) +6NH3(aq)
[yellowish brown]
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[Cu(H2O)6]2+ + 2- OH [Cu(OH)2(H2O)4] pale blue ppt
[Cu(OH)2(H2O)4] pale blue ppt + 4NH3
[Cu(NH3)4(H2O)2]+ pale blue solution
I told
you
Excess ammonia
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Reagent\Ion and initial
colour
Cr3+(aq)
green
Mn2+(aq) very
pale pink ~ colourless
Fe2+(aq)
pale green
Fe3+(aq)
yellow-brown
Co2+(aq)
pink
Ni2+(aq)
green
Cu2+(aq)
blue
Zn2+(aq)
colourless
Al3+(aq)
colourless
initial NaOH(aq)
strong base/alkali
green gel. ppt. of
Cr(OH)3
white gel. ppt. but
darkens with oxidation
Mn(OH)2
Mn2O3
MnO2
dark green ppt. =>
brown on oxidation
Fe(OH)2
Fe(OH)3
brown gel. ppt. of
Fe(OH)3
blue gel. ppt. that
turns pink on standing
Co(OH)2
green gel. ppt. of
Ni(OH)2
gel. blue ppt. of
Cu(OH)2
white gel. ppt. of
Zn(OH)2
white gel. ppt. of
Al(OH)3
excess NaOH(aq)
strong base/alkali
ppt. dissolves to give clear
green solution of
complex ion
[Cr(OH)6]3-
no further effect - just as above with more oxidation
no further effect - just as above with more oxidation
no further effect - just as above with more oxidation
no further effect
no further effect
no further effect ppt. dissolves -
clear solution,
colourless complex ion
[Zn(OH)4]2-
ppt. dissolves -
clear solution,
colourless complex ion
[Al(OH)6]3-
initial NH3(aq)
weak base/alkali
green gel. ppt. of
Cr(OH)3
white ppt. darkens with
oxidation from O2
Mn(OH)2 Mn2O3 MnO2
dark green ppt. turns brown -
oxidation
Fe(OH)2
Fe(OH)3
brown gel. ppt. of
Fe(OH)3
blue gel. ppt. that turns pink on
standing
Co(OH)2
green gel. ppt. of
Ni(OH)2
gel. blue ppt. of
Cu(OH)2
white gel. ppt. of
Zn(OH)2
white gel. ppt. of
Al(OH)3
excess NH3(aq)
weak base/alkali
dissolves - clear green solution of
complex ion
[Cr(NH3)6]3+
no further effect
no further effect
no further effect
ppt. dissolves - clear brown solution of
complex ion
[Co(NH3)6]2+
dissolves - clear pale
blue solution of complex
ion
[Ni(NH3)6]2+
ppt. dissolves to give clear blue
solution of complex ion
[Cu(NH3)4(H2O)2]2+
ppt. dissolves -
clear colourless solution of
[Zn(NH3)4]2+
no further effect
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Which of the following reagents would
enable you to separate iron(III)
hydroxide from a mixture of iron(III)
hydroxide and copper(II) hydroxide?
A Dilute hydrochloric acid
B Aqueous ammonia
C Dilute nitric acid
D Sodium hydroxide solution
Jeelaan
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Splitting of 3d orbital Placing ligands around a central ion causes the
energies of the d orbitals to changeSome of the d orbitals gain energy and some lose
energyIn an octahedral complex, two (z2 and x2-y2) go
higher and three go lowerIn a tetrahedral complex, three (xy, xz and yz) go
higher and two go lowerDegree of splitting depends on the CENTRAL ION and the LIGANDThe energy difference between the levels affects
how much energy is absorbed when an electron is promoted. The amount of energy governs the colour of light absorbed.
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3d 3d
OCTAHEDRAL TETRAHEDRAL
Splitting of 3d orbital
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increasing energy
Cu2+ (‘free ion’) 3d energy level
degenerate
[Cu(H2O)6]2+
3d level split by water ligands
ΔEenergy
difference
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increasing energy
Cu2+ (‘free ion’) 3d energy level
degenerate
[Cu(H2O)6]2+
3d level split by water ligands
ΔEenergy
difference
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Coloured Ions
Absorbed colour nm Observed colour nmVIOLET 400 GREEN-YELLOW 560BLUE 450 YELLOW 600BLUE-GREEN 490 RED 620YELLOW-GREEN 570 VIOLET 410YELLOW 580 DARK BLUE 430ORANGE 600 BLUE 450RED 650 GREEN 520
The observed colour of a solution depends on the wavelengths absorbed
Copper sulphate solution appears blue because the energy absorbed corresponds to red and yellow
wavelengths. Wavelengths corresponding to blue light aren’t absorbed.
ENERGY CORRESPONDING TO THESE COLOURS IS ABSORBED
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The observed colour of a solution depends on the wavelengths absorbed
white lightblue and green not absorbed
WHITE LIGHT GOES IN
SOLUTION APPEARS BLUE
Coloured Ions
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a solution of copper(II)sulphate is blue becausered and yellow wavelengths are absorbed
Coloured Ions
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a solution of copper(II)sulphate is blue becausered and yellow wavelengths are absorbed
Coloured Ions
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a solution of nickel(II)sulphate is green becauseviolet, blue and red wavelengths are absorbed
Coloured Ions
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Observed colourAbsorbed colour
More on Colours
The colour observed in a compound/ complex is actually the complementary colour of the colour absorbed
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Hemantha Welihena
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Colorful cations
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Colorful cations
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A hydrated transition metal ion is colourless. Which
of the following could be the electronic
configuration of this ion?
A [Ar] 3d54s2
B [Ar] 3d8
C [Ar] 3d104s2
D [Ar] 3d10
Nuha
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Copper(II) sulfate solution is blue. This is because
A excited electrons emit light in the blue region of the spectrum as they drop back to the ground state.B excited electrons emit light in the red region of the spectrum as they drop back to the ground state.C electrons absorb light in the red region of the spectrum and the residual frequencies are observed.D electrons absorb light in the blue region of the spectrum and the residual frequencies are observed.
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Shahudhaan
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Uses of Transition Elements
• Chromium’s name comes from the Greek word for color, chrome.
• Many other transition elements combine to form substances with brilliant colors.
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Transition metals are any of various metallic elements such as chromium, iron and nickel that have valence electrons in two shells instead of only one.
Ag + Cu Ag + Cu A valence electron refers to
a single electron that is responsible for the chemical properties of the atom.
Polychromic sun glasses
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+ + 2+
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An important anti-cancer drug. It is a square planar,4 co-ordinate complex of platinum.
Cis-platin
Chemotherapy drugs
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Intercalating agents wedge between bases along the DNA. The intercalated drug molecules affect the structure of the DNA, preventing polymerase and other DNA binding proteins from functioning properly. The result is prevention of DNA synthesis, inhibition of transcription and induction of mutations.
Cis-platin
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Transition metal ions can be identified by color.
Sc Ti V Cr Mn Fe Co Ni Cu Cu Zn
+3 colourless
+3 violet
+3 blue
+3 ruby
+2 pale pink
+2 pale green
+2 pink
+2 green
+ 2 colourless
+ 1 colourless
+2 colourless
+5 yellow
+6 yellow or orange1
+7 purple
+3 red-brown
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Colours of complex ions
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Electronic configuration Ion Colour
Ti [Ar]4s23d2 Ti3+[Ar]4s03d1
Ti4+[Ar]4s03d0 colourless
V [Ar]4s23d3 V2+[Ar]4s03d3 Violet
V3+[Ar]4s03d2 Green
V4+[Ar]4s03d1 Blue
V5+[Ar]4s03d0 Yellow
Cr [Ar]4s13d5 Cr2+[Ar]4s03d4
Cr3+[Ar]4s03d3
Cr6+[Ar]4s03d0
Aqua ComplexesWhen in solution, the ions of Transition metals form what are known as aqua complexes. These are complex ions with water as the ligands. The different ions form complexes with different colours, as shown in the table below:
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Electronic
configuration
Ion Colour
Mn [Ar]4s23d5 Mn2+[Ar]4s03d5 Pale pink
Mn4+[Ar]4s03d3
Mn6+[Ar]4s03d1
Mn7+[Ar]4s03d0
Fe [Ar]4s23d6 Fe2+[Ar]4s03d6 Pale green
Fe3+[Ar]4s03d5 Orange, yellow or brown
Co [Ar]4s23d7 Co2+[Ar]4s03d7 Deep pink
Ni [Ar]4s23d8 Ni2+[Ar]4s03d8 Pale green
Cu [Ar]4s13d10 Cu+[Ar]4s03d10
Cu2+[Ar]4s03d9 Blue
Zn [Ar]4s23d10 colourless Hemantha Welihena
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Cu2+, Cr2+, Co2+, Ni2+
[Cr(H2O)6]2+, [Cu(H2O)6]2+
[Cu(H2O)4]2+, [Cu(NH3)4]2+
[Cu(H2O)2(NH3)4]2+, [Ni(NH3)6]2+
[Ni(NH3)4]2+, [Co(H2O)6]2+
VO2
Blue colour ions
A solution containing the [CoCl4]2-(aq) ion
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Transition Metal Complexes Contents• Aqueous metal ions• Acidity of hexaaqua ions• Introduction to the reactions of complexes• Reactions of cobalt• Reactions of copper• Reactions chromium• Reactions on manganese• Reactions of iron(II)• Reactions of iron(III)• Reactions of silver and vanadium• Reactions of aluminium
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The aqueous chemistry of ions-Hydrolysiswhen salts dissolve in water the ions are stabilisedthis is because water molecules are polarhydrolysis can occur and the resulting solution can
become acidicthe acidity of the resulting solution depends on the
cation presentthe greater the charge density of the cation, the more
acidic the solution
cation charge ionic radius reaction with water pH of chloride Na 1+ 0.095 nm Mg 2+ 0.065 nm Al 3+ 0.050 nm the greater charge density of the cation...
the greater the polarising power and the more acidic the solution Hemantha Welihena
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when salts dissolve in water the ions are stabilisedthis is because water molecules are polarhydrolysis can occur and the resulting solution can become acidicthe acidity of the resulting solution depends on the cation presentthe greater the charge density of the cation, the more acidic the solution
cation charge ionic radius reaction with water / pH of chloride
Na 1+ 0.095 nm dissolves 7 Mg 2+ 0.065 nm slight hydrolysis less than 7 Al 3+ 0.050 nm vigorous hydrolysis less than 7
the greater charge density of the cation...
the greater the polarising power andthe more acidic the solution
Hemantha Welihena
The aqueous chemistry of ions-Hydrolysis
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Complex Ions- LigandsFormation ligands form co-ordinate bonds to a central transition metal ion
Ligands atoms, or ions, which possess lone pairs of electronsform co-ordinate bonds to the central iondonate a lone pair into vacant orbitals on the central species
Ligand Formula Name of ligandchloride Cl¯ chlorocyanide NC¯ cyanohydroxide ¯OH hydroxooxide O2- oxowater H2O aqua
ammonia NH3 ammine
some ligands attach themselves using two or more lone pairsclassified by the number of lone pairs they usemultidentate and bidentate ligands lead to more stable complexes
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some ligands attach themselves using two or more lone pairsclassified by the number of lone pairs they usemultidentate and bidentate ligands lead to more stable complexes
Unidentate form one co-ordinate bond Cl¯, ¯OH, ¯CN, NH3, and H2O
Bidentate form two co-ordinate bonds H2NCH2CH2NH2 , C2O42-
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Complex Ions- Ligands
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some ligands attach themselves using two or more lone pairsclassified by the number of lone pairs they usemultidentate and bidentate ligands lead to more stable complexes
Multidentate form several co-ordinate bonds
EDTA
An important complexing agent
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Complex Ions- Ligands
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[Cu(H2O)6]2+ [CuCl4]2- [Cu(H2NCH2CH2NH2)3]2+ [Cu(edta)]2+
Increasing stability
[Cu(H2O)6]2+ + 4Cl- [CuCl4]
2- + 6H2O
[CuCl4]2- + 3H2NCH2CH2NH2 [Cu(H2NCH2CH2NH2)3]2+ + 4Cl-
[Cu(H2O)6]2+ + edta4- [Cu(edta)]2- + 6H2O
Relative stability of complex ions
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Hemantha Welihena
Complex Ions- Ligands
+2
+2
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ISOMERISATION IN COMPLEXES
GEOMETRICAL (CIS-TRANS) ISOMERISM
Square planar complexes of the form [MA2B2]n+ exist in two forms
trans platin cis platin
An important anti-cancer drug. It is a square planar, 4 co-ordinate complex of platinum.
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The transition metal complex Pt(NH3)2Cl2 exists as two geometric isomers. This is because the complex
A is square-planar.B is tetrahedral.C contains a double bond.D is octahedral
Which of the following species is not able to act as a ligand in the formation of transition metal complexes?
A C6H5NH2
B NH3
C NH2CH2CH2CH2NH2
D NH4
Jeelan
Hemantha Welihena
Shafeeu
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Co-ordination number and Shape the shape of a complex is governed by the number of ligands around the central ion the co-ordination number gives the number of ligands around the central ion a change of ligand can affect the co-ordination number
Co-ordination No. Shape Example(s)
6 Octahedral [Cu(H2O)6]2+
4 Tetrahedral [CrCl4]-
4 Square planar Pt(NH3)2Cl2
2 Linear [Ag(NH3)2]+
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The coordination number of the central metal atom or ion in a complex is the number of dative covalent bonds formed by the central metal atom or ion in a complex.
ComplexThe central metal atom or
ion in the complexCoordination
number
[Ag(NH3)2]+ Ag+ 2
[Cu(NH3)4]2+ Cu2+ 4
[Fe(CN)6]3– Fe3+ 6
Co-ordination number
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Linear [CuCl2]-
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The Aqueous Chemistry of IonsTheory aqueous metal ions attract water molecules
many have six water molecules surrounding themthese are known as hexaaqua ionsthey are octahedral in shapewater acts as a Lewis Base – a lone pair donorwater forms a co-ordinate bond to the metal ionmetal ions accept the lone pair - Lewis Acids
Theory aqueous metal ions attract water moleculesmany have six water molecules surrounding themthese are known as hexaaqua ionsthey are octahedral in shapewater acts as a Lewis Base – a lone pair donorwater forms a co-ordinate bond to the metal ionmetal ions accept the lone pair - Lewis Acids
Acidity as charge density increases, the cation has a greater attraction for waterthe attraction extends to the shared pair of electrons in water’s O-H bonds the electron pair is pulled towards the O, making the bond more polarthis makes the H more acidic (more d+)it can then be removed by solvent water molecules to form H3O+(aq).
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Hydrolysis- Equations , Deprotination
M2+ ions [M(H2O)6]2+(aq) + H2O(l) [M(H2O)5(OH)]+(aq) + H3O+(aq)
the resulting solution will now be acidic as there are more protons in the water
this reaction is known as hydrolysis - the water causes the substance to split up
Stronger bases (e.g. CO32- , NH3 and ¯OH ) can remove further protons...
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M2+ ions [M(H2O)6]2+(aq) + H2O(l) [M(H2O)5(OH)]+(aq) + H3O+(aq)
the resulting solution will now be acidic as there are more protons in the water
this reaction is known as hydrolysis - the water causes the substance to split up
Stronger bases (e.g. CO32- , NH3 and ¯OH ) can remove further protons...
Hemantha Welihena
Hydrolysis- Equations , Deprotination
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M3+ ions [M(H2O)6]3+(aq) + H2O(l) [M(H2O)5(OH)]2+(aq) + H3O+(aq)
the resulting solution will also be acidic as there are more protons in the water
this SOLUTION IS MORE ACIDIC due to the greater charge density of 3+ ions
Stronger bases (e.g. CO32- , NH3 and ¯OH ) can remove further protons...
Hemantha Welihena
Hydrolysis- Equations , Deprotination
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Lewis bases can attack the co-ordinated water molecules. Theoretically, a proton can be removed from each water molecule turning the water from a neutral molecule to a negatively charged hydroxide ion. This affects the overall charge on the complex ion.
[M(H2O)6]2+(aq) [M(OH)(H2O)5]+(aq) [M(OH)2(H2O)4](s)
[M(OH)2(H2O)4](s) [M(OH)3(H2O)3]¯(aq) [M(OH)4(H2O)2]2-(aq)
[M(OH)4(H2O)2]2-(aq) [M(OH)5(H2O)]3-(aq) [M(OH)6]4-(aq)
When sufficient protons have been removed the complex becomesneutral and precipitation of a hydroxide or carbonate occurs.
e.g. M2+ ions [M(H2O)4(OH)2](s) or M(OH)2
M3+ ions [M(H2O)3(OH)3](s) or M(OH)3
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Hydrolysis of Hexaaquaions
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Lewis bases can attack the co-ordinated water molecules. Theoretically, a proton can be removed from each water molecule turning the water from a neutral molecule to a negatively charged hydroxide ion. This affects the overall charge on the complex ion.
[M(H2O)6]2+(aq) [M(OH)(H2O)5]+(aq) [M(OH)2(H2O)4](s)
[M(OH)2(H2O)4](s) [M(OH)3(H2O)3]¯(aq) [M(OH)4(H2O)2]2-(aq)
[M(OH)4(H2O)2]2-(aq) [M(OH)5(H2O)]3-(aq) [M(OH)6]4-(aq)
AMPHOTERIC CHARACTERMetal ions of 3+ charge have a high charge density and their hydroxides can dissolve in both acid and alkali.
[M(H2O)6]3+(aq) [M(OH)3(H2O)3](s) [M(OH)6]3-(aq)
¯ OH
H+
¯ OH
H+
Precipitated
¯ OH
H+Insoluble SolubleSoluble
¯ OH
H+
¯ OH
H+
OH
H+
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Hydrolysis of Hexaaquaions
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The examples aim to show typical properties of transition metals and their compounds.
One typical properties of transition elements is their ability to form complex ions.Complex ions consist of a central metal ion surrounded by co-ordinated ions or molecules known as ligands. This can lead to changes in ...
• colour • co-ordination number• shape • stability to oxidation or reduction
Reactiontypes ACID-BASE
LIGAND SUBSTITUTION
PRECIPITATION
REDOX
A-B
LS
OX
Ppt
RED REDOX
Hemantha Welihena
Types of Reaction
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The examples aim to show typical properties of transition metals and their compounds.
LOOK FOR...
substitution reactions of complex ions
variation in oxidation state of transition metals
the effect of ligands on co-ordination number and shape
increased acidity of M3+ over M2+ due to the increased charge density
differences in reactivity of M3+ and M2+ ions with ¯OH and NH3
the reason why M3+ ions don’t form carbonates
amphoteric character in some metal hydroxides (Al3+ and Cr3+)
the effect a ligand has on the stability of a particular oxidation state
Hemantha Welihena
Types of Reaction
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• aqueous solutions contain the pink, octahedral hexaaquacobalt(II) ion
• hexaaqua ions can also be present in solid samples of the hydrated salts
• as a 2+ ion, the solutions are weakly acidic but protons can be removed by bases...
¯ OH [Co(H2O)6]2+(aq) + 2¯OH (aq) [Co(OH)2(H2O)4](s) + 2H2O(l)
pink, octahedral blue / pink ppt. soluble in XS NaOH
ALL hexaaqua ions precipitate a hydroxide with ¯OH (aq).
Some re-dissolve in excess NaOH
A-B
Hemantha Welihena
Reaction of Cobalt(II)
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NH3 [Co(H2O)6]2+(aq) + 2NH3(aq) [Co(OH)2(H2O)4](s) + 2NH4+(aq)
ALL hexaaqua ions precipitate a hydroxide with NH3 (aq). It removes protons
A-B
Hemantha Welihena
Reaction of Cobalt(II)
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NH3 [Co(H2O)6]2+(aq) + 2NH3(aq) [Co(OH)2(H2O)4](s) + 2NH4+(aq)
ALL hexaaqua ions precipitate a hydroxide with NH3 (aq). It removes protons
Some hydroxides redissolve in excess NH3(aq) as ammonia substitutes as a ligand.
[Co(OH)2(H2O)4](s) + 6NH3(aq) [Co(NH3)6]2+(aq) + 4H2O(l) + 2¯OH (aq)
A-B
LS
Hemantha Welihena
Reaction of Cobalt(II)
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NH3 [Co(H2O)6]2+(aq) + 2NH3(aq) [Co(OH)2(H2O)4](s) + 2NH4+(aq)
ALL hexaaqua ions precipitate a hydroxide with NH3 (aq). It removes protons
Some hydroxides redissolve in excess NH3(aq) as ammonia substitutes as a ligand.
[Co(OH)2(H2O)4](s) + 6NH3(aq) [Co(NH3)6]2+(aq) + 4H2O(l) + 2¯OH (aq)
but ... ammonia ligands make the Co(II) state unstable. Air oxidises Co(II) to Co(III)
[Co(NH3)6]2+(aq) [Co(NH3)6]3+(aq) + e¯
yellow / brown octahedral red / brown octahedral
A-B
LS
OX
Hemantha Welihena
Reaction of Cobalt(II)
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CO32- [Co(H2O)6]2+(aq) + CO3
2-(aq) CoCO3(s) + 6H2O(l)
mauve ppt.
Hexaaqua ions of metals with charge 2+ precipitate a carbonate butheaxaaqua ions with a 3+ charge don’t.
Cl¯ [Co(H2O)6]2+(aq) + 4Cl¯(aq) [CoCl4]2-(aq) + 6H2O(l)
pink, octahedral blue, tetrahedral
• Cl¯ ligands are larger than H2O
• Cl¯ ligands are negatively charged - H2O ligands are neutral
• the complex is more stable if tetrahedral - less repulsion between ligands
• adding excess water reverses the reaction
LS
Ppt
Hemantha Welihena
Reaction of Cobalt(II)
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Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).
¯ OH [Cu(H2O)6]2+(aq) + 2¯OH (aq) [Cu(OH)2(H2O)4](s) + 2H2O(l)
blue, octahedral pale blue ppt. insoluble in XS NaOH
A-B
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Reaction of Cobalt(II)
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One method of manufacturing hydrazine (N2H4) involves the action of sodium chlorate(I) on excess ammonia at 443 K and 50 atm. The yield is normally around 80% but, if just 1 part per million of copper(II) ions is present, the yield drops to 30%.
The most likely explanation for this is the ability of copper(II) ions to
A form complex ions with ammonia.B catalyse reactions producing other nitrogen
compounds.C reduce the hydrazine as it is formed.D reduce the sodium chlorate.
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Arusham
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Reactions between sodium chlorate(I) on excess ammonia
NaOCl(aq) + 2NH3(g) N2H4(g) + NaCl(aq) + H2O(l)
2NaOCl(aq) + 2NH3(aq) 2NH2Cl(g) + 2NaOH(aq)
2NaOCl(aq) + 2NH3(aq) 2NaONH3(aq) + Cl2(g)
3NaOCl(aq) + NH3(aq) 3NaOH(aq) + NCl3(g)
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Metal English name of ion Latin name of ion
V3+ Vanadium(iii) ion Vanadate(iii) ion
Cr3+ Chromium(iii) ion Chromate(iii) ion
Mn2+ Manganese(ii) ion Manganate(ii) ion
Fe2+ Iron(ii) ion Ferrate(ii) ion
Co2+ Cobalt(ii) ion Cobaltate(ii) ion
Ni2+ Nickel(ii) ion Nickalate(ii) ion
Cu2+ Copper(ii) ion Cuprate(ii) ion
Zn2+ Zinc(ii) ion Zinicate(ii) ion
Ag+ Silver(i) ion Argentate(i) ion
Pt2+ Platinum(ii) ion Platinate(ii) ion[V(H2O)6]3 [Cr(H2O)2]2
hexaaquavanadium(ii) hexaaquachromate(ii) Hemantha Welihena
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LigandType of
complexExample Name of Complex
Water Aqua- [Cr(H2O)6]3+ hexaaquachromium(III) ion
Ammonia Ammine- [Ag(NH3)2]+ diamminesilver(I) ion
Hydroxide ion Hydroxo- [Zn(OH)4]2- tetrahydroxozincate(lI) ion
Chloride ion Chloro- [CuCl4]2- tetrachlorocuprate(lI) ion
Cyanide ion Cyano- [Fe(CN)6]3- hexacyanoferrate(III) ion
Nitrite ion Nitro- [CO(NO2)6]3- hexanitrocobaltate(III) ion
Carbon monoxide
Carbonyl- Ni(CO)4 tetracarbonylnickel( 0)
Ethane-l,2-diamine
Ethane-I,2-diamine-
[Cr(en)3]3+ tris(ethane-l,2-diamine)chromium(III) ion
edta edta- [Zn(edta)]2- edtazincate(lI) ion
Cl-, NH3 Mixed [CoCl2(NH3)4]+ tetraamminedichlorocobalt(III) ion-OH, H2O Mixed [Fe(OH)2(H2O)4]+ tetraaquadihydroxoiron(III) ion
Alfred Stock nomenclature
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Nomenclature of Transition Metal
Complexes with Monodentate
LigandsIUPAC conventions
1. (a) For any ionic compound the cation is named before the anion
(b) If the complex is neutral the name of the complex is the name of the compound
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1. (c) In naming a complex (which may be neutral, a cation or an anion)
the ligands are named before the central metal atom or ion
the liqands are named in alphabetical order (prefixes not
counted)
(d) The number of each type of ligands are specified by the Greek prefixes1 mono- 2 di 3 tri
4 tetra- 5 penta- 6 hexa-
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1. (e) The oxidation number of the metal ion in the complex is indicated immediately after the name of the metal using Roman numerals
[CrCl2(H2O)4]Cltetraaquadichlorochromium(III) chloride
[CoCl3(NH3)3]triamminetrichlorocobalt(III)
K3[Fe(CN)6]potassium hexacyanoferrate(III)
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2. (a) The root names of anionic ligands
always end in “-o”
– CN cyano
Cl–
chloro
Br
bromo
I
iodo
– OH hydroxo
NO2 nitro
SO42
sulphato
H
hydrido(b) The names of neutral ligands are the names of the molecules
except NH3, H2O, CO and NO
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3. (a) If the complex is anionic
the suffix “-ate” is added to the end of the name of the metal,
followed by the oxidation number of that metal
tetrachlorocuprate(II) ion[CuCl4]2–
hexacyanoferrate(III) ion[Fe(CN)6]3
tetrachlorocobaltate(II) ion[CoCl4]2
Name of the complexFormula
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(a) Write the names of the following compounds.
(i) [Fe(H2O)6]Cl2
(ii) [Cu(NH3)4]Cl2
(iii) [PtCl4(NH3)2]
(iv) K2[CoCl4]
(v) [Cr(NH3)4SO4]NO3
(vi) [Co(H2O)2(NH3)3Cl]Cl
(vii) K3[AlF6]
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Hexaaquairon(II) chloride
Tetraamminecopper(II) chloride
Diamminetetrachloroplatinum(IV)
Potassium tetrachlorocobaltate(II)
Tetraamminesulphatochromium(III) nitrate
(i) [Fe(H2O)6]Cl2
(ii) [Cu(NH3)4]Cl2
(iii) [PtCl4(NH3)2]
(iv) K2[CoCl4]
(v) [Cr(NH3)4SO4]NO3
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(a) (vi) [Co(H2O)2(NH3)3Cl]Cl
triamminediaquachlorocobalt(II) chloride
(vii) K3[AlF6]
potassium hexafluoroaluminate
Al has a fixed oxidation state (+3)
no need to indicate the oxidation state
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(b) Write the formulae of the following compounds.
(i) pentaamminechlorocobalt(III) chloride
(ii) Ammonium hexachlorotitanate(IV)
(iii) Tetraaquadihydroxoiron(II)
[Co(NH3)5Cl]Cl2
(NH4)2[TiCl6]
[Fe(H2O)4(OH)2]
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The colours of many gemstones are due to the presence of small quantities of d-block metal ions
Coloured Ions
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Reactions of Copper (II)Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).
¯ OH [Cu(H2O)6]2+(aq) + 2¯OH (aq) [Cu(OH)2(H2O)4](s) + 2H2O(l)
blue, octahedral pale blue ppt. insoluble in XS NaOH
NH3 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(OH)2(H2O)4](s) + 2NH4+(aq)
blue ppt. soluble in excess NH3
then [Cu(OH)2(H2O)4](s) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2¯OH (aq)
royal blue
NOTE THE FORMULA
A-B
A-B
LS
Hemantha Welihena
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Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).
¯ OH [Cu(H2O)6]2+(aq) + 2¯OH (aq) [Cu(OH)2(H2O)4](s) + 2H2O(l)
blue, octahedral pale blue ppt. insoluble in XS NaOH
NH3 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(OH)2(H2O)4](s) + 2NH4+(aq)
blue ppt. soluble in excess NH3
then [Cu(OH)2(H2O)4](s) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + ¯2OH (aq)
deep blue
NOTE THE FORMULA
CO32- [Cu(H2O)6]2+(aq) + CO3
2-(aq) CuCO3(s) + 6H2O(l)
blue ppt.
A-B
A-B
LS
Ppt
acidbase
acid base
baseacid
Reactions of Copper (II)
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Cl¯ [Cu(H2O)6]2+(aq) + 4Cl¯(aq) [CuCl4]2-(aq) + 6H2O(l)
yellow, tetrahedral
• Cl¯ ligands are larger than H2O and are charged
• the complex is more stable if the shape changes to tetrahedral
• adding excess water reverses the reaction
2Cu2+(aq) + 4I¯(aq) 2CuI(s) + I2(aq)
off - white ppt.
• a redox reaction• used in the volumetric analysis of copper using sodium thiosulphate
LS
REDOX
Reactions of Copper (II)
Hemantha Welihena
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The aqueous chemistry of copper(I) is unstable compared to copper(0) and copper (II).
Cu+(aq) + e¯ Cu(s) E° = + 0.52 V
Cu2+(aq) + e¯ Cu+(aq) E° = + 0.15 V
subtracting 2Cu+(aq) Cu(s) + Cu2+(aq) E° = + 0.37 V
This is an example of DISPROPORTIONATION where one species is simultaneously oxidised and reduced to more stable forms. This explains why the aqueous chemistry of copper(I) is very limited.
Copper(I) can be stabilised by formation of complexes.
Reactions of Copper (II)
Disproportionation Reactions
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Formula of copper(II) complex Colour of the complex
[Cu(H2O)4]2+
[CuCl4]2–
[Cu(NH3)4]2+
[Cu(H2NCH2CH2NH2)]2+
[Cu(EDTA)]2–
Pale blue
Yellow
Deep blue Violet
Sky blue
Colours of some copper(II) complexes
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This question concerns the chemistry of copper. In the sequence below, A, B, C, D, E and F all contain copper in various oxidation states.
Hemantha Welihena
Cu(OH)2(H2O)4 CuO
[Cu(NH3)4(H2O)2]
Cu [Cu(H2O)6] Cu(NH3)2+2+
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(a) Identify, by name (including the oxidation state where appropriate) or formula,the copper-containing species in the sequence.A ............................................ B CuO ............................................C .............................................D .............................................E .............................................F .............................................
Hemantha Welihena
Cu(OH)2(H2O)4
[Cu(NH3)4(H2O)2]Cu
[Cu(H2O)6]2+
[Cu(NH3)]2+
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(b) Identify, by name or formula, the reagent that would be used to convert B into CuSO4(aq)........................................................................................(c) (i) C and F are the same type of chemical species. Name this type........................................................................................(ii) Explain why C is coloured but F is colourless.Copper ion in C has partially filled d orbitalCopper ion in F has (completely) filled d orbitalsElectronic transitions between partially filled (d) orbitals (of
different energy) are possible*(iii) Explain why F changes into C on shaking.Copper(I) is oxidized (to copper(II))
Hemantha Welihena
H2SO4(aq)
complex(es)
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(d) The reaction of copper(I) iodide to form D and E is a disproportionation.(i) Explain the term disproportionation.(ii) .............................................................................................
.............................................................................................(ii) Write an ionic equation for this reaction. State symbols are not required.(iii) Use the relevant standard reduction (electrode) potentials, from the table on page 17 of your data booklet, to calculate the E cell value for this reaction,giving your answer with the appropriate sign.*(iv) If copper(I) iodide is treated with nitric acid, rather than sulfuric acid, a blue solution is still formed but no pink solid. Use the standard electrode potentials on page 15 of your data booklet to explain this. Quote any data that you use...........................................................................................................................................................................................................
Hemantha Welihena
Simultaneous increase or decrease in oxidation number
2Cu Cu + Cu+ 2+
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Hemantha Welihena
(a) A = copper(II) hydroxide / Cu(OH)2
[Cu(OH)2(H2O)4]
B = copper(II) oxide / CuO
C = tetraamminecopper(II) / [Cu(NH3)]4 /[Cu(NH3)4(H2O)2]
[Cu(NH3)6] / hexaamminecopper(II)
D = copper / Cu / copper(0) / Cu(0)
E = copper(II) sulfate / CuSO4 / Cu / [Cu(H2O)6]
F = diamminecopper(I) / [Cu(NH3)](b) (Dilute) sulfuric acid / H2SO4 / H2SO4(aq)concentrated H2SO4(c)(i) (transition metal / d-block element) complex(es) /complex ion(s)
2+
2+2+
2+
2+
2+
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(c)(ii) Copper ion in C has partially filled d orbital(s) /subshell / 3d9 unpaired d electron d shell Copper ion in F has (completely) filled d orbitals / subshell / 3d2
Electronic transitions between partially filled (d) orbitals (of different energy) are possibleORElectronic transitions between (completely) filled (d) orbitals (of different energy) are not possible(c)(iii) Copper(I) is oxidized (to copper(II)) F / it is oxidized by oxygen / air Second mark(d)(i) (simultaneous) oxidation and reduction OR Simultaneous increase or decrease in oxidation numberof an element(d)(ii) 2Cu+ Cu + Cu2+
OR 2CuI + 2H+ Cu + Cu2+ + 2HIOR 2CuI Cu + Cu2+ + 2I−
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(d)(iii) The use of cell notation (as in the Data Booklet SEP
table) in place of equations
e.g. Cu+(aq) | Cu(s) E = +0.52 (V) (from the data book the
equations are) Cu+(aq) + e− Cu(s) E = +0.52 (V)
Cu2+(aq) + e− Cu+(aq) E = +0.15 (V) So E cell = 0.52 − 0.15 =
+0.37 (V)
(d)(iv) In both schemes the use of cell notation (as in the
Data Booklet SEP table e.g. Cu2+(aq) | Cu(s) E = +0.34 (V)
Relevant half equations are Cu+2(aq)+2e−Cu(s) E = +0.34 (V)
2NO3−(aq) + 4H+(aq)+2e−N2O4(g)+2H2O(l) E = +0.80 (V)
OR NO3−(aq)+3H+(aq)+2eHNO2(aq)+H2O(l) E = +0.94 (V)
Correct overall equation scores both marks:
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Cu + 2 NO3−+ 4H+ Cu2+ + N2O4 + 2H2O
OR Cu + NO3−+ 3H+ Cu2+ + HNO2 + H2O
So Eocell is +0.46 (V) (or +0.60 (V) or just ‘positive’)
Scheme 2 (oxidation of copper(I)
Cu2+(aq) + e−Cu+(aq) E = +0.15 (V)
2NO3−(aq) +4H+(aq) + 2e− N2O4(g) + 2H2O(l)E = +0.80 (V)
OR NO3−(aq) + 3H+(aq) + 2e−HNO2(aq) + H2O(l) E = +0.94 (V)
Correct overall equation
2Cu+ + 2NO3− +4H+ 2Cu2+ + N2O4 + 2H2O
2Cu+ + NO3−+3H+ 2Cu2+ + HNO2+ H2O
So Ecell is +0.65 (V) (or +0.79 (V)
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Reactions of Chromium(III)Chromium(III) ions are typical of M3+ ions in this block.
Aqueous solutions contain the violet, octahedral hexaaquachromium(III) ion.
As an acid [Cr(H2O)6]3+(aq) + 3¯OH (aq) [Cr(OH)3(H2O)3](s) + 3H2O(l)
violet, octahedral green ppt. soluble in XS NaOH
As with all hydroxides the precipitate reacts with acid
As a base [Cr(OH)3(H2O)3](s) + 3H+(aq) [Cr(H2O)6]3+(aq)
being a 3+ hydroxide it is AMPHOTERIC as it dissolves in excess alkali
[Cr(OH)3(H2O)3](s) + 3¯OH (aq) [Cr(OH)6]3-(aq) + 3H2O(l)
green, octahedral
A-B
A-B
A-B
Am
ph
oteric n
ature
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With EXCESS SODIUM HYDROXIDE, the precipitate redissolves
[Cr(H2O) 3(OH) 3] + 3¯OH [Cr(OH)6]3- + 3H2O
Am
ph
oteric n
ature
green ppt. soluble in XS NaOH
green, octahedral
Hemantha Welihena
[Cr(H2O) 3(OH) 3] + 3H [Cr(H2O)6]+3 + 3H2O+
green ppt. soluble in XS HCl
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CO32- 2 [Cr(H2O)6]3+(aq) + 3CO3
2-(aq) 2[Cr(OH)3(H2O)3](s) + 3H2O(l) + 3CO2(g)
The carbonate is not precipitated but the hydroxide is.
high charge density of M3+ makes the solution too acidic to form the carbonate
CARBON DIOXIDE IS EVOLVED.
NH3 [Cr(H2O)6]3+(aq) + 3NH3(aq) [Cr(OH)3(H2O)3](s) + 3NH4+(aq)
green ppt. soluble in XS NH3
With EXCESS AMMONIA, the precipitate redissolves
[Cr(OH)3(H2O)3](s) + 6NH3(aq) [Cr(NH3)6]3+(aq) + 3H2O(l) + 3¯OH(aq)
Reactions of Chromium(III)
Hemantha Welihena
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Oxidation In the presence of alkali, Cr(III) is unstable and can be oxidised to Cr(VI)
2Cr3+(aq) + 3H2O2(l) + 10¯OH(aq) 2CrO42-(aq) + 8H2O(l)
green yellow
Acidification of the yellow chromate will produce the orange dichromate(VI) ion
Reduction Chromium(III) can be reduced to the less stable chromium(II) by zinc in acid
2 [Cr(H2O)6]3+(aq) + Zn(s) 2 [Cr(H2O)6]2+(aq) + Zn2+(aq)
green blue
OX
RED
Reactions of Chromium(III)
Hemantha Welihena
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Occurrence dichromate (VI) Cr2O72- orange
chromate (VI) CrO42- yellow
Interconversion dichromate is stable in acid solution
chromate is stable in alkaline solution
in alkali Cr2O72-(aq) + 2¯OH(aq) 2CrO4
2-(aq) + H2O(l)
in acid 2CrO42-(aq) + 2H+(aq) Cr2O7
2-(aq) + H2O(l)
Reactions of Chromium(VI)
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Being in the highest oxidation state (+6), chromium(VI) will be an oxidising agent.
In acidic solution, dichromate is widely used in both organic (oxidation of alcohols) and inorganic chemistry.
It can also be used as a volumetric reagent but with special indicators as its colour change (orange to green) makes the end point hard to observe.
Cr2O72-(aq) + 14H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l) [ E° = +1.33 V ]
orange green
• Its E° value is lower than that of Cl2 (1.36V) so can be used in the presence of Cl¯ ions
• MnO4¯ (E° = 1.52V) oxidises chloride in HCl so must be acidified with sulphuric acid
• chromium(VI) can be reduced back to chromium(III) using zinc in acid solution
Oxidation Reactions of Chromium(VI)
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This is then oxidised by warming it with hydrogen
peroxide solution. You eventually get a bright yellow
solution containing chromate(VI) ions
wow
Hemantha Welihena
Oxidation Reactions of Chromium(VI)
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Stabilizing an unusual oxidation number:chromium(II) ethanoate, Cr2(CH3CO2)4(H2O)2
Chromium(II) acetate hydrate, also known as chromous acetate, is the coordination compound with the formula Cr2(CH3CO2)4(H2O)2. This formula is commonly abbreviated Cr2(OAc)4(H2O)2. This red-coloured compound features a quadruple bond .The preparation of chromous acetate once was a standard test of the synthetic skills of students due to its sensitivity to air and the dramatic colour changes that accompany its oxidation. It exists as the dihydrate and the anhydrous forms.
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Background
Chromium(II) ethanoate is interesting because it
gives an example of the way in which the
formation of a complex can sometimes stabilise an
oxidation number which would otherwise be
unstable. Chromium(II) ions are normally very
readily oxidised to chromium(III) by the oxygen
of the air. Solutions containing Cr (aq) are green
and those containing Cr (aq) are blue, whereas
chromium (II) ethanoate is red
3+
2+
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Chromium(II) acetate (aqueous solution)
Chromium(II) acetate hydrate, also known as chromous acetate, is the coordination compound with the formula Cr2(CH3CO2)4(H2O)2. This formula is commonly abbreviated Cr2(OAc)4(H2O)2. This red-coloured compound features a quadruple bond .The preparation of chromous acetate once was a standard test of the synthetic skills of students due to its sensitivity to air and the dramatic colour changes that accompany its oxidation. It exists as the dihydrate and the anhydrous forms.
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The apparatus shown in the procedure section can be used, or one which will perform similarly.Zinc in the presence of acid reduces chromium(VI) to chromium(II). Excess zinc also reacts with hydrogen to producehydrogen gas. The build-up of pressure in the apparatus forces the solution into the complex reagent.
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Method1 Dissolve 1 g of sodium dichromate(VI) in 5 cm of water, and put it in the 50 cm round bottomed flask.2 Add 3 g of zinc in equal proportions by mass of powder and granulated.3 Put a mixture of 20 cm concentrated hydrochloric acid and 10 cm water in the tap funnel. Put 10 cm of saturated sodium ethanoate solution in the boiling tube. The solubility of sodium ethanoate in water is about 30% by mass.4 Set up the apparatus as shown in the diagram.5 Add the hydrochloric acid to the mixture in the flask and leave the tap funnel partially open to allow the hydrogen which is generated to escape. The reduction of the chromium soon reaches a green stage and gradually becomes increasingly blue over 10–20 minutes.
33
3
3
3
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6 When the solution is distinctly blue, and while hydrogen is still being generated, close the tap onthe funnel. The pressure of the hydrogen will force the blue solution over into the saturated sodium ethanoate. A red precipitate of chromium(II) ethanoate should be formed, and the solution will contain dissolved red chromium(II) ethanoate.7 Dismantle the apparatus and pour the remaining blue solution containing Cr (aq) into another boiling tube. Keep the two boiling tubes unstoppered, side by side in a rack.
2+
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Throughout this experiment, you must wear eye protection.
The chemicals and procedures used in this experiment are extremely hazardous, so you must take even more care than usual to reduce risks from them by using suitable control measures.
Solid sodium dichromate(VI) is very toxic and oxidising and is classed as a category 2 carcinogen. It is an irritant to all tissues and you must wear gloves when handling this solid. Avoid inhaling any tiny crystals. Zinc powder is highly flammable.
Concentrated hydrochloric acid is corrosive. The hydrogen evolved is extremely flammable.
As hydrogen is evolved, naked flames must be kept clear.
Safety
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PreparationAn aqueous solution of a Cr(III) compound is first reduced to the chromous state using zinc as a reductant.The resulting blue chromous solution is treated with sodium acetate. Immediately chromous acetate precipitates as a bright red powder.
Cr6+ + 2Zn → Cr2+ + 2Zn2+ 2 Cr2+ + 4 OAc- + 2 H2O → Cr2(OAc)4(H2O)2
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The synthesis of Cr2(OAc)4(H2O)2 has been
traditionally used to test the synthetic skills and patience of inorganic laboratory students in universities because the accidental introduction of a small amount of air into the apparatus is readily indicated by the discoloration of the otherwise bright red product.An alternative route to related chromium(II) carboxylates starts with chromocene :
4 HO2CR + 2 Cr(C5H5)2 → Cr2(O2CR)4 + 4 C5H6
The advantage to this method is that it provides anhydrous derivatives.
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Because it is so easily prepared, Cr2(OAc)4(H2O)2 is often used as a starting
material for other, chromium(II) compounds. Also many analogues have been prepared using other carboxylic acids in place of acetate and using different bases in place of the water.
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SafetyThroughout this experiment, you must wear eye protection. The chemicals and procedures used in this experiment are extremely hazardous, so you must take even more care than usual to reduce risks from them by using suitable control measures. Solid sodium dichromate(VI) is very toxic and oxidising and is classed as a category 2 carcinogen. It is an irritant to all tissues and you must wear gloves when handling this solid. Avoid inhaling any tiny crystals. Zinc powder is highly flammable. Concentrated hydrochloric acid is corrosive. The hydrogen evolved is extremely flammable. As hydrogen is evolved, naked flames must be kept clear.
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Questions
1 Show, using electrode potentials from pages 14–16
in the Edexcel Data Booklet, that zinc should reduce
chromium from 16 in sodium dichromate(VI) to 12 in
chromium(II) ethanoate.
2 What happens to the colour of the Cr (aq) solution
when it is allowed to stand open to the air?
3 Does the colour of the chromium(II) ethanoate
solution also change over the same period of time?
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Four reactions involving the transition elements
copper and chromium are given below.
1 Cu2+(aq) + 2-OH(s) Cu(OH)2
2 [Cu(H2O)4(OH)2](s) + 4NH3(aq) [Cu(NH3)4]2+(aq) +
2-OH(aq) + 2H2O(l)
3 [Cr(H2O)3(OH)3](s) +3-OH(aq) [Cr(OH)6](aq) +3H2O(l)
4 [Cr(H2O)3(OH)3](s) + 3H+(aq) [Cr(H2O)6]3+(aq)
(a) Which reaction produces a dark blue solution?
A 1 B 2 C 3 D 4
Dheema
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(b) Which two reactions show the amphoteric behaviour of a metal hydroxide?A 1 and 2
B 2 and 3C 2 and 4
D 3 and 4
(c) Predict, without calculation, which reaction has the most negative value for ΔSsystem.A 1
B 2C 3
D 4
Nashaya
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In the reaction of manganate(VII) ions with
reducing agents in strongly acidic solution,
the half-reaction for the reduction is
A MnO4 + 4H+ + 3e– MnO2 + 2H2O
B MnO4 + 4H+ + 5e– Mn2+ + 2H2O
C MnO4 + 8H+ + 3e– Mn2+ + 4H2O
D MnO4 + 8H+ + 5e– Mn2+ + 4H2O
Zinal
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When a solution containing 0.10 mol of
chromium(III) chloride, CrCl3.6H2O, is treated
with excess silver nitrate solution, 0.20 mol of
silver chloride, AgCl, is immediately precipitated.
The formula of the complex ion in the solution is
A [Cr(OH)6]3-
B [Cr(H2O)6]3+
C [CrCl(H2O)5]2+
D [CrCl2(H2O)4]2+
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When concentrated ammonia solution is added to a green solution of chromium(III) sulfate, a green precipitate is formed which slowly dissolves in excess of the concentrated ammonia solution.
The chromium-containing species formed in these reactions are
Green precipitate Resulting solution
A Cr(OH)3 [Cr(OH)6]
B Cr(OH)3 [Cr(NH3)6]
C (NH4)2CrO4 [Cr(OH)6]
D (NH4)2CrO4 [Cr(NH3)6]
Hemantha WelihenaJaah
3-
3-
3+
3+
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1 Four complex ions have the following formulae:A Cu(edta)2-
B Zn(H2O)62+
C Ni(NH3)62+
D CrCl42-
(a) Which complex ion is most likely to be tetrahedral in shape?
A B C D(b) Which complex ion is most likely not to be coloured?A B C D(c) Each of these complex ions may be formed by ligand exchange from an aqua complex. For which complex ion is the entropy change of this reaction most positive?A B C D
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Platinum forms a complex with the formula Pt(NH3)2Cl2 and chromium forms a complex ion with the formula CrCl4 .
(a) Considering the shapes of these complexes, A both complexes are square planar. B both complexes are tetrahedral. C Pt(NH3)2Cl2 is tetrahedral and CrCl4 is square planar. D Pt(NH3)2Cl2 is square planar and CrCl4 is tetrahedral.
(b) Considering the structures of these complexes, A both complexes form stereoisomers. B neither complex forms a stereoisomer. C Pt(NH3)2Cl2 forms a stereoisomer but CrCl4 does not. D CrCl4 forms a stereoisomer but Pt(NH3)2Cl2 does not.
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Asleefa
Haneefa
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(c) Considering the bonding between the central atom and the ligands in these complexes, A the bonding in both complexes is dative covalent. B the bonding in both complexes is ionic. C the bonding in Pt(NH3)2Cl2 is dative covalent and in CrCl4 is ionic. D the bonding in Pt(NH3)2Cl2 is ionic and in CrCl4 is dative covalent.
Hemantha Welihena
Zinal
Cr(H2O)63+ + 4Cl CrCl4 + 6H2O
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Manganese
show oxidation states of +2, +3, +4 ,+5, +6 and +7 in its compounds
Variable Oxidation States of
Manganese and their
Interconversions
The most common oxidation states
+2, +4 and +7
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IonOxidation state of
manganese in the ionColour
Mn2+
Mn(OH)3
Mn3+
MnO2
MnO43
MnO42–
MnO4–
+2
+3
+3
+4
+5
+6
+7
Very pale pink
Dark brown
Red
Black
Bright blue
Green
Purple
Colours of compounds or ions of manganese in different oxidation states
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(a)
Colours of compounds or ions of manganese in differernt oxidation states: (a) +2; (b) +3; (c) +4
(b) (c)
Mn2+(aq) Mn(OH)3(aq) MnO2(s)
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(e)(d)
Colours of compounds or ions of manganese in differernt oxidation states: (d) +6; (e) +7
MnO42–(aq) MnO4
–(aq)
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Reactions of Manganese (VII) in its highest oxidation state therefore Mn(VII) will be an
oxidising agent
occurs in the purple, tetraoxomanganate(VII)
(permanganate) ion (MnO4¯)
acts as an oxidising agent in acidic or alkaline solutionacidic MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l) E° = + 1.52 V
N.B. Acidify with dilute H2SO4 NOT dilute HCl
alkaline MnO4¯(aq) +2H2O(l) +3e¯ MnO2(s) +4¯OH(aq) E° = + 0.59 V
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Some unusual oxidation states
You will probably have seen manganese in its common oxidation states during your study of chemistry. Manganese(II) sulfate occurs as very pale pink crystals in the hydrated form. Manganese(IV) oxide is a black powder which is often used as a catalyst. Potassium manganate(VII) occurs as very dark purple crystals and forms a purple aqueous solution, which is a powerful oxidizing agent.
You are less likely to have seen compounds containing the other oxidation states of manganese, which are manganese(VI), manganese(V), manganese(III) and manganese(I).
However, compounds containing each of these four oxidation states can be prepared.
MnSO4, MnO2 , KMnO4 , Mn , Mn , Mn , Mn6+ 5+ 3+ +
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Manganese(VI)
Manganese(VI), in MnO4, can be prepared in a reverse
disproportionation reaction, by reacting manganate(VII) ions
with manganese(IV) oxide in alkali.Equation 12MnO4(aq) +MnO2(s) +4OH 3MnO4(aq)+2H2O(l) Ecell = –0.03V
The reaction is not thermodynamically favourable under standard conditions. However, the Ecell value can be made positive by increasing the concentration of hydroxide ions so that green manganate(VI) ions form.Manganese(V)Manganese(V) can be formed by adding a little potassium manganate(VII) to very concentrated (12 mol dm3-) aqueous sodium hydroxide. The solution slowly becomes blue as manganate(V) ions, MnO3(aq), form. The ionic half-equations are:
- - 2-
2-
-
Ɵ
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Equation 2
MnO4(aq) + H2O(l) +2e MnO3aq) + 2OH(aq)Equation 3 4OH(aq) 2H2O(l) + O2(g) + 4eManganese(III)
A deep red solution containing manganese(III) ions is formed by the oxidation of manganese(II) hydroxide by potassium manganate(VII) in acid solution. The ionic equation for the reaction is:Equation 4
MnO4(aq) + 4Mn(OH)2(s) + 16H+(aq) 5Mn3+(aq)+ 12H2O(l)
Manganese(I)
Manganese(I) ions are not stable in aqueous solution, but do form stable complex ions. They can be made by reducing hexacyanomanganate(II) ions, Mn(CN)64-, tohexacyanomanganate(I) ions, Mn(CN)65- .
-
-
-
-
-
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(a) (i) Give the formula of manganese(IV) oxide.
(ii) How do catalysts speed up reactions?
(iii) Explain how transition metal ions can act as homogeneous catalysts.
(b) (i) Suggest why the preparation of manganate(VI) ions, MnO4, in equation 1,may be described as a reverse disproportionation reaction by considering the relevant oxidation states.
(ii) The two half-equations which are combined to form equation 1 are
MnO4(aq) + e MnO42-(aq) E = +0.56 V
MnO42-(aq) +2H2O(l) +2e MnO2(s)+4OH(aq) E =+0.59 VExplain, by reference to these half-equations, why increasing
the concentration of hydroxide ions alters the electrode potential to make the preparation of manganate(VI) ions more likely.
-
- -
- Ɵ
Ɵ
2-
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(c) Use equations 2 and 3 to answer the following questions.(i) Identify the gas formed in the preparation of manganate(V) ions.(ii) By appropriately combining these two equations, write the ionic equation for the formation of manganate(V) ions from manganate(VII) ions.(iii) Identify the main hazard and state how you would minimize the associated risk in this preparation of manganate(V) ions.(d) Identify the reagents you would use to make manganese(II) hydroxide for the preparation of manganese(III) ions.(e) (i) Draw a dot and cross diagram to show the electron arrangement in the cyanide ion, CN.(ii) Explain how the cyanide ion acts as a ligand.(iii) Suggest the name of the shape of the hexacyanomanganate(I) ion.
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15(a)(i) MnO2
15(a)(ii) • They provide alternative routes/mechanisms forreactions• With lower activation energies/Ea OR catalysts loweractivation energy /Ea• So a greater proportion of/more particles/reactants havesufficient energy/Ea (to react)/greater frequency of/more successful collisions15(a)(iii) Transition metals form various/variable oxidation states. They are able to donate and receive electrons/they are able to oxidize and reduce/they are able to be oxidized and reduced/ions contain partially filled(sub-)shells of d electrons. Transition metals form various/variable oxidation states.• They are able to donate and receive electrons/they are able
to oxidize and reduce/they are able to be oxidized and reduced/ions contain partially filled (sub-)shells of d electrons
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15(b)(i) Two (less stable) oxidation states/one higher and one lower oxidation state of the same/an element react to form one(more stable) oxidation state
15(b)(ii) When the hydroxide ion concentration is increased) the equilibrium (of the second half equation) moves to the left/back• E becomes less positive/more negative/decreases/reduces• Therefore Ecell becomes positive (so reaction feasible) 15(c)
(i) Oxygen/oxygen gas
15(c)(iii) (Hazard –) the sodium hydroxide/alkali iscorrosive/caustic/burns (skin)/attacks the skinOR attacks the cornea/eye/causes blindness 15(d) Manganese(II)/manganous sulfate(solution) manganese(II) salt – chloride,bromide, iodide, nitrateSodium hydroxide (solution) ,soluble hydroxide
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15(e)(i)
15(e)(ii) The non-bonding / lone pair of electrons on the carbon forms a dative covalent/coordinate bond (to central metal ion)
15(e)(iii) Octahedral/octahedron
Mn+
- CN
- CN- CN
- CNNC -
NC -
C N
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The compounds of vanadium, vanadium
oxidation states of +2, +3, +4 and +5
forms ions of different oxidation states
show distinctive colours in aqueous solutions
Variable Oxidation States of Vanadium and
their Interconversions
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IonOxidation state of vanadium in the
ion
Colour in aqueous solution
V2+(aq)
V3+(aq)
VO2+(aq)
VO2+(aq)
+2
+3
+4
+5
Violet
Green
Blue
Yellow
Colours of aqueous ions of vanadium of different oxidation states
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V
H2O
H2OH2O
H2O
OH2
OH2
2+
V
H2O
H2OH2O
H2O
OH2
OH2
3+
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In an acidic medium
the vanadium(V) state usually occurs in
the form of VO2+(aq) dioxovanadium(V) ion
the vanadium(IV) state occurs in the form
of VO2+(aq) oxovanadium(IV) ion
Variable Oxidation States of Vanadium and
their Interconversions
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In an alkaline medium
the stable form of the vanadium(V) state is
VO3–(aq), metavanadate(V) or
VO43–(aq), orthovanadate(V),
in strongly alkaline medium
Variable Oxidation States of Vanadium and
their Interconversions
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Compounds with vanadium in its highest oxidation state (i.e. +5)
strong oxidizing agents
Variable Oxidation States of Vanadium and
their Interconversions
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Vanadium of its lowest oxidation state(i.e. +2)
in the form of V2+(aq)
strong reducing agent
easily oxidized when exposed to air
Variable Oxidation States of Vanadium and
their Interconversions
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The most convenient starting material
ammonium metavanadate(V) (NH4VO3)
a white solid
the oxidation state of vanadium is +5
Interconversions of the common oxidation states of vanadium can be carried out readily in the laboratory
Variable Oxidation States of Vanadium and
their Interconversions
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Interconversions of Vanadium(V) species
VO2+(aq) V2O5(s) VO3
(aq) VO43(aq)
OH
H+
OH
H+
OH
H+
Yellow orange yellow colourless
Vanadium(V) can exist as cation as well as anion
Variable Oxidation States of Vanadium and
their Interconversions
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Interconversions of Vanadium(V) species
VO2+(aq) V2O5(s) VO3
(aq) VO43(aq)
OH
H+
OH
H+
OH
H+
Yellow orange yellow colourless
In acidic medium In alkaline mediumAmphoteric
Variable Oxidation States of Vanadium and
their Interconversions
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Interconversions of Vanadium(V) species
VO2+(aq) V2O5(s) VO3
(aq) VO43(aq)
OH
H+
OH
H+
OH
H+
Yellow orange yellow colourless
In acidic medium In alkaline mediumAmphoteric
Give the equation for the conversion : V2O5 VO2+
V2O5(s) + 2H+(aq) 2VO2+(aq) + H2O(l)
Variable Oxidation States of Vanadium and
their Interconversions
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Interconversions of Vanadium(V) species
VO2+(aq) V2O5(s) VO3
(aq) VO43(aq)
OH
H+
OH
H+
OH
H+
Yellow orange yellow colourless
In acidic medium In alkaline mediumAmphoteric
Give the equation for the conversion : V2O5 VO3
V2O5(s) + 2OH(aq) 2VO3(aq) + H2O(l)
Variable Oxidation States of Vanadium and
their Interconversions
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Interconversions of Vanadium(V) species
VO2+(aq) V2O5(s) VO3
(aq) VO43(aq)
OH
H+
OH
H+
OH
H+
Yellow orange yellow colourless
In acidic medium In alkaline medium
Give the equation for the conversion : VO3 VO2
+
VO3(aq) + 2H+(aq) VO2
+(aq) + H2O(l)
Amphoteric
Variable Oxidation States of Vanadium and
their Interconversions
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V5+
H
O
H
H
O
H
H
O
H
H
O
H
VO43(aq) + 8H3O+
8H2O
O
H
H
V5+ ions does not exist in water since it undergoes vigorous hydrolysis to give VO4
3
The reaction is favoured in highly alkaline solution
orthovanadate(V) ion
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V VO43(aq) orthovanadate(V) ion
Cr CrO42(aq) chromate(VI) ion
Mn MnO4(aq) manganate(VII) ion
Draw the structures of VO43, CrO4
2 and MnO4
O
Cr
OO-
O-
O
Mn
OO
O-
O
V-O
O-O-
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V5+
H
O
H
H
O
H
H
O
H
H
O
H
VO3(aq) + 6H3O+
6H2O
O
H
H
The reaction is favoured in alkaline solution
VO3 is a polymeric anion like SiO3
2
Metavanadate(V) ion
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Metavanadate(V) ion, (VO3)nn
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V5+
H
O
H
H
O
H
H
O
H
H
O
H
VO2+(aq) + 4H3O+
4H2O
O
H
H
The reaction is favoured in acidic solution
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The action of zinc powder and concentrated hydrochloric acid
vanadium(V) ions can be reduced sequentially to vanadium(II) ions
Variable Oxidation States of Vanadium and
their Interconversions
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VO2+(aq)
yellow
Zn
conc. HCl
VO2+(aq) blue
Zn
conc. HCl
V3+(aq) green
Zn
conc. HCl
V2+(aq)
violet
Variable Oxidation States of Vanadium and
their Interconversions
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(a)
Colours of aqueous solutions of compounds containing vanadium in four different oxidation states:(a) +5; (b) +4; (c) +3; (d) +2
(b) (c) (d)
VO2+(aq) VO2+(aq) V3+(aq) V2+(aq)
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The feasibility of the changes in oxidation state of vanadium
can be predicted using standard electrode potentials
Half reaction (V)
Zn2+(aq) + 2e– Zn(s)
VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l)
VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l)
V3+(aq) + e– V2+(aq)
–0.76
+1.00
+0.34
–0.26
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Under standard conditions
zinc can reduce
1. VO2+(aq) to VO2+(aq)> 0
> 0
> 02. VO2+(aq) to V3+(aq)
3. V3+(aq) to V2+(aq)
Variable Oxidation States of Vanadium and
their Interconversions
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2 × (VO2+(aq) + 2H+(aq) + e–
VO2+(aq) + H2O(l)) = +1.00 V–) Zn2+(aq) + 2e– Zn(s) = –0.76 V
2VO2+(aq) + Zn(s) + 4H+(aq)
2VO2+(aq) + Zn2+(aq) + 2H2O(l)
= +1.76 V
Variable Oxidation States of Vanadium and
their Interconversions
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2 × (VO2+(aq) + 2H+(aq) + e–
V3+(aq) + H2O(l)) = +0.34 V
–) Zn2+(aq) + 2e– Zn(s) = –0.76 V
2VO2+(aq) + Zn(s) + 4H+(aq)2V3+(aq) + Zn2+(aq) +
2H2O(l)
= +1.10 V
Variable Oxidation States of Vanadium and
their Interconversions
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2 × (V3+(aq) + e– V2+(aq)) =
–0.26 V–) Zn2+(aq) + 2e– Zn(s) = –0.76 V
2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+
(aq)
= +0.50 V
Variable Oxidation States of Vanadium and
their Interconversions
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REACTIONS OF VANADIUM
Reduction using zinc in acidic solution shows the various oxidation states of vanadium.
Vanadium(V) VO2+(aq) + 2H+(aq)+ e¯ VO2+(aq)+ H2O(l)
yellow blue
Vanadium(IV) VO2+(aq) + 2H+(aq) +e¯ V3+(aq) + H2O(l)
blue blue/green
Vanadium(III) V3+(aq) + e¯ V2+(aq)
blue/green lavender
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Redox chemistry of vanadium V , Z = 23 , 1s22s22p63s23p63d34s2
NH4VO3(s) + NaOH(aq) NH3(g) + H2O + NaVO3
white
VO\ 3(aq) + 2H+(aq) VO+2(aq) + H2O(l) white yellow
VO2+ VO2
+ & VO2+ VO2+ V3+ V2+
+5 +5 and +4 +4 +3 +2
yellow green= yellow +blue
blue green lavender =violet
VO2+ VO2
+ & VO2+ VO2+ V3+ V2+
dioxovanadium(V)ion Oxovanadium(IV)ion Vanadium(III)ion Vanadium(III)ion
Zn Zn2+ + e-
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+5 +5 and +4 +4 +3 +2
yellow green= yellow
+blue
blue green lavender =violet
VO2+ VO2
+ & VO2+ VO2+ V3+ V2+
dioxovanadium(V)ion
Oxovanadium(IV)ion
Vanadium(III)ion
Vanadium(III)ion
COLOURFUL COMPOUNDSRedox chemistry of vanadium V , Z = 23 , 1s22s22p63s23p63d34s2
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Dioxovanadium(V) ions reduce by zinc in acidic solution
with zinc in HCl:
2VO2+(aq) + 4H+(aq) + 2e – 2VO2+ (aq) + 2H2O (l)
Zn(s) Zn2+ (aq) + 2e –
2VO2+(aq) + 4H+(aq) + Zn(s) 2VO2+(aq) +Zn2+(aq)+2H2O (l)
dioxovanadium(V)ion
oxovanadium(IV)ion
+4 +3 with zinc in HCl
2VO2+ + 2e – + 4H+(aq) 2V3+(aq) + 2H2O (l)
Zn(s) Zn2+(aq) + 2e –
2VO2+ + Zn(s) + 4H+(aq) 2V3+(aq)+ Zn2+(aq) +2H2O (l)
[V(H2O)6]3+ :
hexaaquavanadium(iii)ion
Oxovanadium(IV) ions reduce by zinc in acidic solution
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+3 +2 with zinc in HCl [V(H2O)6]2+:
hexaaquavanadium(ii)ion2V3+(aq) + 2e – 2V2+ (aq)
Zn(s) Zn2+(aq) + 2e –
2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+ (aq)
Vanadium(III) ions reduce by zinc in acidic solution
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Zn Zn2+ + e-
+5 +5 and +4 +4 +3 +2
yellow green= yellow +blue
blue green lavender =violet
VO2+ VO2
+ & VO2+ VO2+ V3+ V2+
dioxovanadium(V)ion Oxovanadium(IV)ion Vanadium(III)ion Vanadium(III)ion
VO2+ VO2
+ & VO2+ VO2+ V3+ V2+
VO\ 3(aq) + 2H+(aq) VO+2(aq) + H2O(l) white yellow
NH4VO3(s) + NaOH(aq) NH3(g) + H2O + NaVO3
white
Dioxovanadium(V) ions reduce by zinc in acidic solution
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The first stage of the series of reductionsLet's look at the first stage of the reduction - from VO2
+ to VO2+. The redox potential for the vanadium half-reaction is given by:
The corresponding equilibrium for the zinc is:
Eɵcell = Eɵ Eɵ = + 1.00 – (-0.76 ) = + 1.76 V RHS LHS
Reaction is feasible
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The other stages of the reactionHere are the E° values for all the steps of the reduction from vanadium(V) to vanadium(II):
…… and here is the zinc value again:
Eɵ / V
Zn2+/ Zn - 0.76
V3+/ V2+ - 0.26
VO2+/ V3+ + 0.34
VO2+ / VO2+ + 1.00
1.101.76
0.50
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If zinc is over. What are the additional things can you observe?
A. If excess hydrogen ions present in the mixture?
B. If nitrate ions present in the mixture?
Re-oxidation of the vanadium(II)
The vanadium(II) oxidation state is easily oxidised back to
vanadium(III) - or even higher oxidation by hydrogen ions
Eɵcell = Eɵ – Eɵ = 0.00 - ( - 0.26) = +0.26V RHS LHS
Reaction is feasible
That means that the vanadium(II) ions will be oxidised to
vanadium(III) ions, and the hydrogen ions reduced to hydrogen.
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Will the oxidation go any further - for example,
to the vanadium(IV) state?
Have a look at the E° values and decide:
Eɵcell = Eɵ – Eɵ = +0.34V –(+0.00V) = +0.34V RHS LHS
Reaction is feasible
That means that the vanadium(III) ions will be oxidised to vanadium(IV) ions, and the hydrogen ions reduced to hydrogen.
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Oxidation by nitric acid
In a similar sort of way, you can work out how far nitric acid will oxidize the vanadium(II).
Here's the first step:
Eɵcell = Eɵ – Eɵ = 0.96V – (- 0.26V) = + 1.22V RHS LHS
Reaction is feasible
Nitric acid will oxidize vanadium(II) to vanadium(III).
The second stage involves these E° values:
Eɵcell = Eɵ – Eɵ = 0.96 – (+ 0.34) = + 0.62V RHS LHS
Reaction is feasible
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Nitric acid will certainly oxidize vanadium(III) to
vanadium(IV).
Will it go all the way to vanadium(V)?
It has got a less positive value, and so the reaction happen.
Eɵcell = Eɵ – Eɵ = 1.00V – (+0.96V ) = +0.04 V RHS LHS
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Zn+2(aq) + 2e- Zn(s) ; E Ɵ = -0.76V
VO2+(aq) + 2H+(aq) + e- H2O(l) + V3+(aq) EƟ = +0.32V
VO2+(aq) + 2H+(aq) + e- H2O(l) + VO2+(aq)
EƟ = + 1.00V
V3+(aq) + e- V2+(aq) EƟ = -0.26
Sn+2(aq) + 2e- Sn(s) ; E Ɵ = -0.14V
EƟ/ V
0.00
Oxidation number
NO3-(aq) + 4H+(aq) + 3e- 2H2O(l) + NO(aq)
E Ɵ= + 0.96V
Redox Chemistry of Vanadium
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Displacement of Ligands and Relative
Stability of Complex Ions
[Fe(H2O)6]2+(aq) + 6–CN (aq)Hexaaquairon(II) ion
[Fe(CN)6]4–(aq) + 6H2O(l)
Hexacyanoferrate(II) ion
Stronger ligand
Weaker ligand
Reversible reaction
Equilibrium position lies to the right
Kst 1024 mol6 dm18
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[Ni(H2O)6]2+(aq) + 6NH3(aq)Hexaaquanickel(II) ion
[Ni(NH3)6]2+(aq) + 6H2O(l)
Hexaamminenickel(II) ion
Stronger ligand
Weaker ligand
The greater the equilibrium constant,the stronger is the ligand on the LHS andthe more stable is the complex on the RHS
The equilibrium constant is called the stability constant, Kst
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Relative strength of some ligands bonding with copper(II) ions
monodentate
bidentate
multidentate
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Test Observation Inference
(a) Observe the appearanceof A.
Pale greensolid. ………………………………………
(b) Measure the pH of adilute aqueous solutionof A using a pH meter.
The pH is 6.0. The type of reaction that has occurred when A dissolved in water is........................................
(c) Add a few drops ofdilute sodium hydroxidesolution to a solutionof A.
A greenprecipitateforms.
The sodium hydroxide is acting as.......................................................The formula of the green precipitate is………………………………………
(d) (d) Leave a sample ofthe green precipitateformed in (c) to stand inair.
The greenprecipitateturns brownon the surface.
The type of reaction that has occurred is………………………….The formula of the brown precipitate is……………………….
(e) (e) Add excess sodiumhydroxide solution toa sample of the greenprecipitate formed in (c).
The greenprecipitatedoes notdissolve.
……………………………………….
(f) Add barium chloridesolution, BaCl2(aq),acidified withhydrochloric acid, to asolution of A.
A whiteprecipitateforms.
The white precipitate is
The table shows a series of tests carried out on a soluble crystalline compound A, which contains one anion and one cation. For each test, complete the table by filling in the inference column.
Identify compound A by name or formula.
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Extent of reaction
∆Stotal J mol-1K-1
Ecell / v Kc
Goes to completion
More positive than+200
More positive than =0.6
> 1010
Equilibrium with more products than reactants
approx+40
approx +0.1
approx102
Roughly equal amounts of reactants and products
approx 0
approx 0
approx 1
Equilibrium with more reactants than products
approx - 40
approx-0.1
approx 10-2
No reactionMore negative
than-200
More negative than -0.6
> 10-10
Entropy , electromotive force and the extent of a reaction
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Thank you